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Question: If the ratio of specific heat of a gas at constant pressure to that at constant volume is \(\gamma \...

If the ratio of specific heat of a gas at constant pressure to that at constant volume is γ\gamma , the change in internal energy of the mass of gas when the volume changes from V{\text{V}} to 2  V{\text{2}}\;{\text{V}} at constant pressure PP, is
A) Rγ1\dfrac{R}{{\gamma - 1}}
B) PVPV
C) PVγ1\dfrac{{PV}}{{\gamma - 1}}
D) γPVγ1\dfrac{{\gamma PV}}{{\gamma - 1}}

Explanation

Solution

The change in the internal energy of the gas depends on the temperature. When the gas is at constant pressure then the change in internal energy also depends on the specific heat of the gas at constant volume and number of moles. Substituting for the specific heat of gas we can reach a new equation.

Complete step by step answer:
Given the pressure is constant. Therefore the expression for the change in internal energy at constant pressure is given as,
ΔU=nCVΔT..........(1)\Delta U = n{C_V}\Delta T..........\left( 1 \right)
Where nn is the number of moles, CV{C_V} is the specific heat of the gas at constant volume and ΔT\Delta T is the change in temperature.
We have the ratio of specific heat of a gas at constant pressure to that at constant volume is γ\gamma .
Therefore CPCV=γ\dfrac{{{C_P}}}{{{C_V}}} = \gamma
Where, CP{C_P} is the specific heat of the gas at constant pressure, and CV{C_V} is the specific heat of the gas at constant volume.
The above expression can be rearranged to
CPCVCV=γ1\Rightarrow \dfrac{{{C_P} - {C_V}}}{{{C_V}}} = \gamma - 1
And CPCV=R{C_P} - {C_V} = R
Where RR is the gas constant.
Substituting in the above expression.
RCV=γ1\Rightarrow \dfrac{R}{{{C_V}}} = \gamma - 1
CV=Rγ1\Rightarrow {C_V} = \dfrac{R}{{\gamma - 1}}
Substituting this in equation (1)
ΔU=nRγ1ΔT............\Delta U = n\dfrac{R}{{\gamma - 1}}\Delta T............(2)
From the ideal gas equation,
PV=nRTPV = nRT
Where PP is the pressure and VV is the volume.
The equation for the constant pressure and change in temperature will be as,
PV=nRΔT\Rightarrow PV = nR\Delta T
Substitute this in the equation (2)\left( 2 \right) .
ΔU=PVγ1\Rightarrow \Delta U = \dfrac{{PV}}{{\gamma - 1}}
Thus the change in internal energy will be PVγ1\dfrac{{PV}}{{\gamma - 1}}.

Therefore, the answer is option (C).

Note:
We have to note that for ideal gas there is no change in internal energy. But whenever a non-ideal gas has been analyzed the change in internal energy happened due to any change in pressure, volume, or temperature.