WBBSE Class 10 Physical Science Solutions Chapter 2 Behaviour of Gases

Detailed explanations in West Bengal Board Class 10 Physical Science Book Solutions Chapter 2 Behaviour of Gases offer valuable context and analysis.

WBBSE Class 10 Physical Science Chapter 2 Question Answer – Behaviour of Gases

Very Short Answer Type Questions :

Question 1.
Why are gaseous molecules always in incessant motion?
Answer:
Because gaseous molecules have no mutual force of attraction.

Question 2.
What is the cause of pressure of a gas taken in a vessel?
Answer:
Gaseous molecules continuously strike against the walls of the container.

Question 3.
In which direction pressure exerted by a gas in vessel does vary, if at all?
Answer:
Equal in all direction.

Question 4.
What is the effect of supplying heat to a gas at constant volume ?
Answer:
Pressure of the gas increases.

Question 5.
What is the cause of temperature of gas ?
Answer:
Kinetic energy possessed by gaseous molecules.

Question 6.
What is the effect of withdrawal of heat from a gas at constant volume ?
Answer:
Temperature of the gas falls.

Question 7.
What are the two constant quantities in Boyle’s law ?
Answer:
Mass of the gas and temperature.

Question 8.
What is the temperature inCelsius scale corresponding to absolute zero?
Answer:
-273°C

Question 9.
What Is meant by ideal gases?
Answer:
Gases which obey Boyle’s law and Charle’s law at all pressure and temperature.

Question 10.
How does the volume of a gas change for each degree celsius rise of temperature at constant pressure?
Answer:
\(\frac{1}{273}\) of volume of the gas at 0°C increases.

Question 11.
What will happen when a gas-filled balloon heated?
Answer:
The balloon will burst as the supplied heat increases the volume of the gas inside the balloon.

Question 12.
What will be the volume of an ideal gas at absolute zero?
Answer:
The volume will be zero.

Question 13.
At what temperature speed of gas molecules becomes zero?
Answer:
At -273°C or 0 K speed of gas molecule becomes zero.

Question 14.
What are the constant quantities inCharle’s law?
Answer:
In Charle’s law, the constant quantities are mass of the gas and pressure.

Question 15.
What are the variable quantities inCharle’s law?
Answer:
In Charle’s law, the variable quantities are temperature and volume.

Question 16.
What are the variable quantities in Boyle’s law?
Answer:
In Boyle’s law the variable quantities are pressure and volume.

Question 17.
What is the freezing temperature of water in absolute scale?
Answer:
273 K

Question 18.
Give a mathematical expression of Boyle’s law.
Answer:
V × P= constant

Question 19.
Give a mathematical expression ofCharle’s law.
Answer:
[\(\frac{V}{T}\) = constant]

Question 20.
What is the SI unit of pressure?
Answer:
SI unit of pressure is Pascal.

Question 21.
How is atmospheric pressure determined ?
Answer:
Atmospheric pressure is determined by Barometer invented by Torricelli.

Question 22.
How is pressure of a gas determined ?
Answer:
The pressure of a gas is determined by Manometer.

Question 23.
What is absolute zero?
Answer:
The temperature at which the volume of a gas reduces to zero is called absolute zero.

Question 24.
IsCharle’s law applicable for liquids ?
Answer:
Charle’s law is not applicable for liquids.

Question 25.
Write the ideal gas equation for one gram-mole gas.
Answer:
PV = RT

Question 26.
What is the value of absolute temperature in Fahrenheit scale ?
Answer:
(-) 4594° F

Question 27.
What is the relation between absolute scale and celsius scale of temperature?
Answer:
If any temperature is t°C in celsius scale and if is TK in Kelvin scale then
T = t + 273

Question 28.
What is Torr?
Answer:
Torr is the unit of pressure used at high vacuum. 1 Torr = 1 mm of Hg

Question 29.
What is Loschmidt number?
Answer:
The numher of molecules present in 1 ml of gas or vapour at STP is known as Loschmidt number.

Question 30.
What is Boyle’s temperature?
Answer:
The temperature above which a real gas behaves like an ideal gas is called Boyle’s temperature.

Question 31.
Give a main property of a gas.
Answer:
Diffusion is a main property of a gas.

Question 32.
How does rate of diffusion of a gas vary with temperature ?
Answer:
Rate of diffusion of a gas increases with increase in temperature.

Question 33.
State the names of the scientists related to kinetic theory of gases.
Answer:
Kinetic theory of gases was proposed by Bernouli and extended by Clausius, Maxwell, Kronig, Boltzmann etc.

Question 34.
What is collision frequency?
Answer:
The number of collisions made by all gas molecules in a unit volume per second is known as collision frequency.

Question 35.
Define torr ?
Answer:
Pressure exerted by exactly 1 mm of mercury column at 0°C and at standard gravity is termed as 1 torr.

Question 36.
Which state of matter has no definite volume and no definite shape?
Answer:
gaseous state.

Question 37.
What is the value of gas constant in SI unit ?
Answer:
8.314 JK^-1 mol^-1

Question 38. How is the unit ‘L’ related to dm³ ?
Answer:
IL=1 dm³

Question 39.
Name the SI unit of pressure ?
Answer:
Pascal

Question 40.
Under what conditions is Boyle’s law is applicable?
Answer:
At constant mass and at constant temperature.

Question 41.
What is the effect of increase of temperature on vapour of a liquid at constant pressure?
Answer:
With rise of temperature vapour of a liquid increases at constant pressure.

Question 42.
Write down the value of R in L atm mol ^-1 k^-1.
Answer:
0.0821 L atm mol^-1 k^-1

Short Answer Type Questions :

Question 1.
State one point of difference between expansion of gas and expansion of solid.
Answer:
For any range of temperature rise all gases expand equally, but different solids expand by different amount for same range of temperature rise.

Question 2.
Why does not the volume of a real gas become zero at absolute zero?
Answer:
Explanation : Any gas or vapour liquefies much before reaching this low temperature -273°C. Again,Charle’s law is valid for gases, not for liquids. Hence, this situation of zero volume of gas is never attained practically.

Question 3.
Why tyre of an automobile is inflated to a lesser pressure in summer than in winter?
Answer:
Explanation : In summer, the room temperature is much higher as compared to winter.Consequently, the same volume of air will exert greater pressure in summer. Therefore, it is always advisable to inflate the tyre to lesser extent in summer as compared to winter to avoid the bursting of tyres.

Question 4.
Is it possible to cool a gas below absolute zero ?
Answer:
Explanation : No, it is not possible to cool a gas below absolute zero. At absolute zero, the kinetic energy of the gas molecules becomes zero. The molecules come to rest at the bottom of the container.
Therefore, pressure of the gas is also zero. It is therefore, impossible to cool a gas below absolute zero, since there is no heat left to be removed from the gas.

Question 5.
What is effusion of a gas?
Answer:
Effusion: It is the passage of a gas under pressure through a tiny hole in a container. Process of effusion is used to separate the isotopes of an element. During effusion, the lighter isotopes escape first from the pores of the container leaving behind the heavier isotopes.

Question 6.
How does an ideal gas differ from a real gas?
Answer:
Explanation : As ideal gas obeys gas laws or the gas equation P V=n R T all temperatures and pressures whereas a real gas obeys the gas laws or the gas equation only at high temperatures and low pressures. There is no cooling or heating effect observed when an ideal gas expands in vacuum whereas a real gas shows either cooling or heating effect.

Question 7.
Is it possible to keep liquid nitrogen in a sealed cylinder at ordinary temperature like ammonia?
Answer:
Reason : No, it is not possible to keep the liquid nitrogen in a sealed container at ordinary temperature like ammonia because critical temperature of nitrogen is very low whereas the critical temperature of ammonia is lighter than that of ordinary temperature.

Question 8.
Why aerated bottles are kept under water during summer ?
Answer:
Explanation : Areated water bottles contain carbon dioxide gas dissolved in aqueous solution under pressure. In summer, the solubility of the gas in solution decreases with rise in temperature.Consequently, the amount of free gas in bottle increases. This leads to increase in pressure. Thus, the bottle may explode. To avoid explosion, bottles are kept under cold water in summer.

Question 9.
State two characteristics of gases.
Answer:
Characteristics of gases :
(i) A gas has no definite shape or volume. It takes the shape and attains the volume in which it is kept. This tendency of scattering of the gas molecules is called expansion of gas.
(ii) If a few gases which do not react with each other, are kept in a container, these gases mix with each other and a homogeneous mixture is prepared. This property is known as diffusion.

Question 10.
What is normal temperature and pressure (NTP) ?
Answer:
NTP or STP (Standard Temperature and Pressure): It is defined as the pressure exerted by a column of mercury 76 cm in height at 0°C, at 45° latitude and at mean sea-level.

Question 11.
Calculate the value of normal preseure inCGS and SI system.
Answer:
Normal pressure inCGS system : 76 × 13.6 × 980 = 1.013 × 10⁶ dyne / cm² Normal pressure in SI system : 1.013 × 10⁵ N / m²=1.013 × 10⁵ pascal

Question 12.
State Boyle’s law.
Answer:
Boyle’s law (1662) : At constant temperature, the volume of a definite mass of any gas’varies inversely as the pressure of the gas.

Question 13.
Deduce the mathematical expression of Boyle’s law.
Answer:
Mathematical expression of Boyle’s law : If P be the pressure and V the volume of a given mass of a gas, then according to Boyle’s law,
V ∝ \(\frac{1}{P}\) (the mass and the temperature remaining constant)
or, V = \(\frac{K}{P}\) ( K is a proportionality constant)
or, PV = K (constant)
This is the mathematical experession of Boyle’s law.

Question 14.
What will be the nature of graph if P is plotted against V for a given mass of gas at constant temperature?
Answer:
Explanation of graph : On plotting P against volume V fat a given temperature, the graph will be a rectangular hyperbola as shown below.

Question 15.
When a balloon infiates, It seems to violate Boyle’s law. Explain.
Answer:
Explanation : When a rubber balloon inflated with the end of a pump, the pressure inside increases but the volume instead of suffering a decrease in accordance with Boyle’s law increases. This happens because with the introduction of air, the mass of the gas increases resulting an increase in volume. Thus, inflation of a balloon by mouth or pump appears to violate Boyle’s law but it is not so in the true sense.

Question 16.
StateCharle’s law.
Answer:
Charle’s law (1787): At constant pressure, the volume of a given mass of a gas increases or decreases by \(\frac{1}{273}\) of its volume at 0°C for each one degree rise or fall in temperature.

Question 17.
Deduce the mathematical expression ofCharle’s law.
Answer:
Mathematical expression of Charle’s law :
Let V₀ be the volume of adiefnite mass of a gas at 0°C, then the volume of 1°C rise in temperature will be =(V₀+V₀ × \(\frac{1}{273}\))
∴ The volume of t°C rise in temperature will be =V₀ + V₀ × \(\frac{1}{273}\) If V_t represents this volume at t°C then
V_t + V₀+V₀ × \(\frac{t}{273}\) = V₀(1 + \(\frac{t}{273}\))
Similarly, when temperature at the same mass of gas be gradually made to decrease, its volume at -t°C is given by
V_t + V₀ + V₀ × \(\frac{t}{273}\) = V₀(1-\(\frac{t}{273}\))

Question 18.
What will be the nature of graph if V_t is plotted against t for a given mass of gas at constant pressure?
Answer:
Explanation of graph : On plotting V_t against t, at a given pressure the graph will be a straight line. If this straight line is extraploted then the straight line meets at ‘ X ‘-axis which represents the absolute temperature (-273°C).

Question 19.
What is absolute zero? Define absolute scale of temperature.
Answer:
Absolute zero: The lowest possible temperature at which a given mass of gas does not occupy any volume or does not exert any pressure is known as absolute zero.
Absolute scale of temperature : The temperature scale which has been devised by taking -273°C as zero and using the magnitude of each degree as the celsius degree is known as absolute or Kelvin scale of temperature.

Question 20.
What is the alternative form ofCharle’s law?
Answer:
Alternative form ofCharle’s law : Suppose at constant pressure, the volume of a certain mass of gas at 0°C = V₀ and at t₁°C and t₂°C temperatures, the volumes are V₁ and V₂ respectively.

According toCharle’s law,

Alternative statement ofCharle’s law : The volume of a given mass of a gas kept at constant pressure is directly proportional to its absolute (Kelvin) temperature.

Question 21.
State Pressure-Temperature relationship or Gay-Lussac’s law.
Answer:
Pressure-Temperature relationship or Gay-Lussac’s law : At constant volume, the pressure of a given mass of a gas increases by \(\frac{1}{273}\) of its original value at 0°C for every 1°C rise in temperature.

Question 22.
State the mathematical expression of Gay-Lussac’s law.
Answer:
Mathematical expression of Gay-Lussac’s law :

Therefore, Gay. Lussac’s law can also be stated. At constant volume, the pressure of a given mass of a gas is directly proportional to its absolute temperature.

Question 23.
Establish the combined law of Boyle’s andCharle’s laws.
Answer:
Combination of Boyle’s law andCharle’s law :
Let V be the volume of a given mass of gas at pressure P and temperature T (absolute).
From Boyle’s law, V ∝ \(\frac{I}{P}\)
[When temperature and mass of the gas are constant]
FromCharle’s law, V ∝ T
[ When pressure and mass of the gas are constant]
∴ V∝ \(\frac{T}{P}\) when both T and P vary.
or, V = \(\frac{KT}{P}\) or, \(\frac{PV}{T}\) = K [K= constant]
Now, if V₁ is the volume of a gas at pressure P₁ and temperature T₁ and V₂ is the volume of the same amount of gas at pressure P₂ and temperature T₂ then,
\(\frac{P_1 V_1}{T_1}=\frac{P_2 V_2}{T_2}\)

Question 24.
What Is equation of state? What is universal gas constant ?
Answer:
Equation of state: From the combination of Boyle’s andCharle’s laws we get, \(\frac{PV}{T}\) = K
The numerical value of the constant of proportionality (K) depends upon quantity of gas and the units in which volume and pressure are expressed but is totally independent of the nature of the gas.
For 1 mole of gas, the constant is termed as universal gas constant.
∴ \(\frac{PV}{T}\) = R
or, PV = RT
This is known as general gas equation.
For ‘ n ‘ moles of the gas therefore, the gas equation may be written as:
PV = nRT

Question 25.
What is the relation between density and pressure of an ideal gas.
Answer:
According to ideal gas equation PV = nRT = \(\frac{m}{MRT}\) where ‘m’ is the mass of gas and ‘m’ its molecular mass,
P = \(\frac{m}{V}\) \(\frac{RT}{M}\)
or, P = d \(\frac{RT}{M}\) where d = \(\frac{m}{V}\) = density of the gas.
∴ d α P constant temperature.

Question 26.
What is real gas? What is ideal gas ?
Answer:
Real gas : It is a gas which does not obey general gas equation and all other gas laws strictly but tends towards ideality at low pressure and high temperature.
Ideal gas : It is a gas obeys the general gas equation and other gas laws under all conditions of temperature and pressure.

Question 27.
State the values of universal gas constant R in different units.
Answer:
Values of universal gas constant R in different units ;

1. 0.0821 lit atm mol^-1 K^-1
2. 8.314 × 10⁷ erg mol^-1 K^-1
3. 8.314 J mol^– K^-1
4. 1.987 cal mol^-1 K^-1

Broad Answer Type Questions :

Question 1.
What are the characteristic properties of gases ?
Answer:
Characteristic properties of gases :

1. Shape and Volume: Since particles of gases are not held in fixed positions and move freely, gases neither have definite shapes nor definite volumes.
2. Homogeneous nature : All the parts of a gas or a gaseous mixture have similar composition throughout.
3. Density : Due to large separation of molecules, gases have large volume and thus low density.
4. Compressibility : On increasing pressure gases can be readily compressed due to the presence of large empty spaces.
5. Random motion : The molecules or atoms of a gas are in a state constinuous zig-zag motion in all directions.
6. Pressure : Due to their random motion, the molecules of the gas collide on the walls of the container and thus exert certain force on walls of the container. The force per unit area is called the gas pressure.
7. Diffusion : Gases mix (diffuse) with each other freely due to the free movement of their molecules.
8. Liquefaction : On cooling and applying pressure, gases can be liquefied. However, a gas has to be cooled below a certain characteristic temperature called critical temperature before it can be liquefied by the application of pressure alone.

Question 2.
What are the postulates of Kinetic theory of gases?
Answer:
Postulates of Kinetic theory of gases :

1. All gases consist of very large number of tiny particles called molecules that are in constant rapid motion.
2. The gas molecules are perfectly round, very hard and separated by large distances. Their actual volume is thus, negligible as compared to the total volume of the gas.
3. The collisions between the gas molecules are perfectly elastic, i.e. there is no loss of energy during these collisions.
4. The distance between gas molecules being very large, there is no effective force of attraction or repulsion between them.
5. The average kinetic energy of the gas molecules is directly proportional to the absolute temperature of the gas.
6. The gas molecules collide with one another and with the walls of the container. The pressure exerted by a gas is due to the bombardment of its molecules on the walls of the vessel.
7. The gas molecules move freely in all directions. Their speed and direction change continously due to collisions among them. As a result, their motion becomes zig-zag or random.

Brain Storming Questions :
Very Short Answer Type Questions :

Question 1.
What is the significance of S.T.P. or N.T.P?
Answer:
Temperature and pressure affects the volume of gas. In order to compare the volumes of any two gases, they are to be brought under some standard condition of temperature and pressure. For this reason, S.T.P. or N.T.P. are introduced during study of gas.

Question 2.
Why do hot air balloons rise higher than cold air balloons?
Answer:
At higher temperature, a given mass of gas occupies more volume as per charles’ law and is thus dense.Cold air being denser than hot air, raises the balloon to a lesser height.

Question 3.
Why do gas balloons brust at high altitudes?
Answer:
We know that, at high altitudes, the atmospheric pressure on the balloon decreases, resulting in increase in volume as per Boyles’law. So the balloon brusts at high altitudes.

Question 4.
Under what conditions the value of \(\frac{P}{T}\) will be constant irrespective of the value of T ?
Answer:
We know that, for a certain mass of gas at constant volume, \(\frac{P}{T}\) = K (constant). So, the value of \(\frac{P}{T}\) is constant for a fixed mass of gas at constant volume no matter what the value of T is. This is obtained from Gay Lussac’s law.

Question 5.
Carbondioxide is heavler than air, yet it does not form the lower layer of air-why?
Answer:
From the definition of diffusion we know that gases in the air get intermixed. Although CO₂ is heavier than air yet it does not form the lower layer.

Question 6.
Wet air is more penetrating than dry air-why?
Answer:
As wet air being lighter than dry air, so wet air diffuses more readily than dry air.

Question 7.
Name an instrument to measure the pressure of a gas.
Answer:
Manometer is used to measure the pressure of a gas.

Question 8.
What do you mean by STP?
Answer:
STP means standard temperature and pressure. Standard temperature is 0°C. 76 cm mercury pressure or 1 atm. pressure is taken as standard.

Question 9.
Can temperature in Keloin scale be negative?
Answer:
No. starting with the zero reading at absolute zero temperature, temperature in kelvin scale is always positive.

Question 10.
What is the dimension of R ?
Answer:
We know that, universal gas constant, R

Question 11.
On which two factors does the molar volume of a gas depend?
Answer:
Molar volume of a gas depends on pressure and temperature.

Question 12.
What is the Boyle temperature of hydrogen gas?
Answer:
Boyle temperature of hydrogen is -156°C.

Question 13.
Mention an example of diffusion applicable to daily life.
Answer:
Detection of fragrance of perfumes occurs due to gaseous diffusion.

Question 14.
What is the nature of collisions between the molecules of a gas?
Answer:
The collisions between the molecules are perfectly elastic i.e.there is no loss of energy during these collisions.

Question 15.
What is the value of kinetic energy of an Ideal gas at absolute zero temperature ?
Answer:
At absolute zero temperature all the molecular motions cease and hence the kinetic energy of an ideal gas becomes zero.

Question 16.
What is molar volume?
Answer:
The volume occupied by one mole of substance at a given temperature and pressure is called the molar volume of the substance.

Numerical Problems :

(i) T_t=t_c+273
(ii) P₁ V₁=P₂ V₂ (at constant temperature and constant mass of gas)
(iii) V₁=V_{\circ(1+\frac{t{273)
(iv) \(\frac{V_1}{T_1}=\frac{V_2}{T_2}\) (at constant pressure and constant mass of gas)
(v) \(\frac{P_1}{T_1}=\frac{P_2}{T_2}\) (at constant volume and constant mass of gas)
(vi) PV = nRT = \(\frac{m}{MRT}\)

Erample 1:
What will be the reading of 30°C in absolute scale of temperature?
Answer:
The reading of 30°C in absolute scale of temperature
= (273 + 30) K = 303K

Example 2:
What is the value of absolute zero in the celsius scale?Calculate its value in Farhenheit scale.
Answer:
The value of absolute zero in celsius scale = -273°C.
Let the value of the above temperature in Farhenheit scale = X° F.
We know, \(\frac{C}{5}\) = \(\frac{F-32}{9}\)
or, \(\frac{-273}{5}\) = \(\frac{X-32}{9}\)
or, 5 x-160 = -2457 or, 5 x = -2297
or, x = -459.4° F
∴ The temperature in Farhenheit scale = -459.4° F

Example 3 :
27°C or 300 K, which temperautre is higher?
Answer:
27°C = (27 + 273)K = 300 K
So, the two temperatures are equal.

Example 4 :
30°C or 300 K, which temperature is higher ?
Answer:
30°C=(30+273) K=303 K
So, 30°C is higher than 300 K

Example 5:
A weather balloon has a volume of 175 L when filled with hydrogen at a pressure of 1.00 atm.Calculate the volume of the balloon when it rises to a height of 2000 m, where the atmospheric pressure is 0.80 atm. Assume that the temperature is constant.
Answer:
P₁ = 1.00 atm
V₁ = 175 L
P₂ = 0.80 atm
V₂ = ?

According to Boyle’s law
P₁ V₁=P₂ V₂
or, V₂=\(\frac{P_1 V_1}{P_2}\) = \(\frac{1 \times 175}{0.80}\)

Example 6:
A sample of oxygen has a volume of 880 mL and a pressure of 740 torr. What additional pressure is required to reduce the volume to 440 mL.
Answer:
P₁ = 740 torr
V₁ = 880 mL
P₂ = ?
V₂ = 440 mL

According to Boyle’s law
P₁V₁ = P₂ V₂
or, P₂ = \(\frac{P_1 V_1}{V_2}=\frac{740 \times 880}{440}\)
= 218.75 L
∴ Additional pressure required
= (1480-740) torr
= 740 torr

Example 7:
A bottle of volume 9 litres contains a gas at 10 atmospheric pressure. How many bottles each of 1 litre capacity can be filled with the gas at 2 atmospheric pressure, at constant ternerature.
Answer:
P₁ = 10 atm
V₁ = 9 lit
P₂ = 2 atm
V₂ = ?

According to Boyle’s law
P₁ V₁=P₂ V₂
or, V₂=\frac{P₁ V₁{P₂=\frac{10 × 9{2=45 lit
∴ 45 bottles, each of 1 litre volume can be filled.

Example 8 :
A sample of helium has a volume of 520 cm³ at 373 K.Calculate the temperature at which the volume will become 260 cm³. Assume that the pressure is constant.
Answer:
According toCharle’s law
V₁ = 520 cm³
T₁ = 373 K
V₂ = 260 cm³
T₂ = ?
or, = 186.5 K

Example 9:
Find the volume of N₂ at 27°C at a pressure of 760 torr, if it occupies 40 mL at STP.
Answer:
V₁ = 40 mL
T₁ = 373 K
T₂ = (27+373) K=300 K
V₂ = ?

According toCharle’s law

\(\frac{V_1}{T_1}=\frac{V_2}{T_2}\)
or, V₂= \(\frac{V_1 \times T_2}{T_1}\) = \( \frac{40 \times 300}{273}\) = 44 mL

Example 10:
The volume of a certain mass of gas at of 400 K and pressure 202600 Pascal is 2 cubic metres. What would be the volume of the gas at 327°C and 152 cm of mercury pressure?
[given 1.013 × 10⁵ pascal =1 atmospheric pressure]
Answer:
1.013 × 10⁵ Pascal = 76 cm of Hg
202600 Pascal = \(\frac{76 \times 202600}{1.013 \times 10^5}\)
= 125 cm of Hg
T₁ = 327°C = (327 + 273) K = 600K
V₁ = 2 m³
V₂ = ?
T₁ = 400 K

According toCharle’s law

\(\frac{V_1}{T_1}=\frac{V_2}{T_2}\)
or, V₂ = \(\frac{V_1 \times T_2}{T_2}\)
or, V₂ = \(\frac{2 \times 600}{400}\)
∴ V₂ = 3 m³

Example 11:
A steel tank contains carbon dioxide at 27°C and a pressure of 10.0 atm.Calculate the internal gas pressure when the tank and the gas are heated to 100°C.
Answer:
P₁ = 10.0 atm
T₁ = (27 + 273) K = 300 K
P₂ = ?
T₂ = (100 + 273) K = 373 K

According to Gay-Lussac’s law.

\(\frac{P_1}{T_1}=\frac{P_2}{T_2}\)
or, P₂ = \(\frac{P_1 \times T_2}{T_1}\)
or, P₂ = \(\frac{10 \times 373}{300}\) = 12.43 atm.

Example 12 :
An iron cylinder contains helium at a pressure at 250 K Pa at 300 K. The cylinder can withstand a pressure of 1 × 10⁶ Pa. The room in which cylinder is placed catches fire. Predict whether the cylinder will blow up before it melts or not.
(M.P. of the cylinder =1800 K )
Answer:
P₁ = 250 KPa = 250 × 10³ Pa
V₁ = 300 K
P₂ = ?
T₂ = 1800 K

According to Gay-Lussac’s law

\(\frac{P_1}{T_1}=\frac{P_2}{T_2}\)
or, P₂ = \(\frac{P_1 \times T_2}{T_{21}}\)
or, P₂ = \(\frac{250 \times 10^3 \times 1800}{300}\) = 1.5 × 10⁶ Pa

Example 13:
A gas having temperature 0°C is heated so that its pressure and volume are doubled. What will be the final temperature of the gas?
Answer:
Let, P₁ = P (say)
by condition, P₂ = 2 P
again V₁ = V (say)
So, V₁ = 2 V
T₁ = (0 + 273) K = 273 K
T₂ = ?

By combining Boyle’s law andCharle’s law we get

Example 14:
A certain quantity of a gas occupies a volume of 1000 cm³ at 760 mm and 27°C. Find the volume of the gas if the pressure and temperature are 1520 mm and 327°C.
Answer:
P₁ = 760 mm
V₁ = 1000 cm³
T₂ = (273 + 27) K = 300 K
P₂ = 1520 mm
T₂ = (273 + 327) K = 600 K
V₂ = ?

By combining Boyle’s law andCharle’s law we get,

Example 15:
7.0 g of a gas at 300 K and 1 atmospheric pressure occupies a volume of 4.1 littres. What is the molecular mass of gas?
Answer:
w = 7.0 g
R = 0.0821 lit atm mol^-1 K^-1
T = 300 K
M = ?
P = 1 atm
V = 4.1 lit

Applying ideal gas equation,

PV = nRT
PV = \(\frac{w}{M}\)
or, M = \(\frac{wRT}{PV}\)
or, M = \(\frac{7 \times 0.0821 \times 300}{1 \times 4.1}\)
∴ M = 42 g-mol^-1

Example 16:
10 g of oxygen are introduced in a vessel of 5 lit capacity at 27°C.Calculate the pressure of the gas in atmospheres in the container.
Answer:
w = 10 g
M = 32
R = (27+273) K=3000 K
V = 5 lit
R = 0.0821 lit atm mol^-1 K^-1
P = ?

Applying ideal gas equation,

P V =n R T
∴ P V = \(\frac{w}{M}\)
or, P = \(\frac{w R T}{V \times M}\)
∴ P = 1.539 atm

Example 17:
The volume of 1 gram mole of oxygen gas is 22400 mL at 760 mm.Calculate the temperature of the gas (Given R=0.082 L-atm mol{ ^-1 K^-1 )
Answer:
P = 670 mm = 1 atm
V = 22400 mL = 22.4 L
R = 0.082 L – atm mol^-1 K^-1

We know from ideal gas equation for 1 mole of an ideal gas

PV = RT
or, T = \(\frac{P V}{R}\) or, T = \(\frac{1 \times 22.4}{0.082}\)
or, T = 273.17 K
Temperature in celsius scale = 273.17 – 273 = 0.17°C