The molar specific heats of an ideal gas at a constant pressure and volume are denoted by $ {{C}_{p}} $ and $ {{C}_{v}} $ respectively. If $ \gamma =\dfrac{{{C}_{p}}}{{{C}_{v}}} $ and R is the universal gas constant, then $ {{C}_{v}} $ is equal to:
(A) $ \dfrac{1+\gamma }{1-\gamma } $
(B) $ \dfrac{R}{\gamma -1} $
(C) $ \dfrac{(\gamma -1)}{R} $
(D) $ \gamma R $
Answer
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Hint: Specific heat of gas at constant volume ( $ {{C}_{v}} $ )is defined as the amount of heat energy required to raise the temperature of one gram of gas through $ \text{1}{}^\circ \text{C} $ , when volume of the gas kept constant.
Specific heat of a gas at a constant pressure ( $ {{C}_{p}} $ ) is defined as the amount of heat energy required to raise the temperature of one gram of gas through 1°C when pressure is kept constant.
Use the relation $ {{C}_{p}}-{{C}_{v}}\text{ =}R $ , to find the required result.
Complete step by step solution
We have given, Molar specific heat of an ideal gas at constant volume is $ {{C}_{v}} $ .
Molar specific heat of an ideal gas at an specific pressure is $ {{C}_{p}} $ .
Since, the ratio of specific heat of an ideal gas at constant pressure to specific heat of an ideal gas at constant volume is given by $ \gamma $ .
i.e., $ \gamma =\dfrac{{{C}_{p}}}{{{C}_{v}}} $ --------- (1)
We have relation between $ {{C}_{p}} $ and $ {{C}_{v}} $ . : $ {{C}_{p}}-{{C}_{v}}\text{ =}R $ ----------- (2)
Here R is universal gas constant.
Divide eq. (2) and $ {{C}_{v}} $
$ \dfrac{{{C}_{p}}}{{{C}_{v}}}\text{-1=}\dfrac{R}{{{C}_{v}}}\text{ } $
Using eq. (1) $ \text{ }\gamma -1=\dfrac{R}{{{C}_{v}}} $
$ {{C}_{v}}=\dfrac{R}{\gamma -1} $
This is the required result for molar specific heat of an ideal gas at constant volume.
Option (B) is the correct answer.
Note
Since, $ {{C}_{p}} $ is greater than $ {{C}_{v}} $
This is because,
When heat is given to gas at constant volume, it is only used in increasing the internal energy of gas i.e., in raising the temperature and no heat is spent in expansion of gas.
When heat is given to gas at constant pressure, it is used in increasing the internal energy and also in the expansion of gas against external pressure.
Therefore to raise the temperature of 1 mole of a gas through $ \text{1}{}^\circ \text{C} $ or 1k, more heat is required at constant pressure than at constant volume. Hence, $ {{C}_{p}} $ ˃ $ {{C}_{v}}\text{ } $ .
Specific heat of a gas at a constant pressure ( $ {{C}_{p}} $ ) is defined as the amount of heat energy required to raise the temperature of one gram of gas through 1°C when pressure is kept constant.
Use the relation $ {{C}_{p}}-{{C}_{v}}\text{ =}R $ , to find the required result.
Complete step by step solution
We have given, Molar specific heat of an ideal gas at constant volume is $ {{C}_{v}} $ .
Molar specific heat of an ideal gas at an specific pressure is $ {{C}_{p}} $ .
Since, the ratio of specific heat of an ideal gas at constant pressure to specific heat of an ideal gas at constant volume is given by $ \gamma $ .
i.e., $ \gamma =\dfrac{{{C}_{p}}}{{{C}_{v}}} $ --------- (1)
We have relation between $ {{C}_{p}} $ and $ {{C}_{v}} $ . : $ {{C}_{p}}-{{C}_{v}}\text{ =}R $ ----------- (2)
Here R is universal gas constant.
Divide eq. (2) and $ {{C}_{v}} $
$ \dfrac{{{C}_{p}}}{{{C}_{v}}}\text{-1=}\dfrac{R}{{{C}_{v}}}\text{ } $
Using eq. (1) $ \text{ }\gamma -1=\dfrac{R}{{{C}_{v}}} $
$ {{C}_{v}}=\dfrac{R}{\gamma -1} $
This is the required result for molar specific heat of an ideal gas at constant volume.
Option (B) is the correct answer.
Note
Since, $ {{C}_{p}} $ is greater than $ {{C}_{v}} $
This is because,
When heat is given to gas at constant volume, it is only used in increasing the internal energy of gas i.e., in raising the temperature and no heat is spent in expansion of gas.
When heat is given to gas at constant pressure, it is used in increasing the internal energy and also in the expansion of gas against external pressure.
Therefore to raise the temperature of 1 mole of a gas through $ \text{1}{}^\circ \text{C} $ or 1k, more heat is required at constant pressure than at constant volume. Hence, $ {{C}_{p}} $ ˃ $ {{C}_{v}}\text{ } $ .
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