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Question: A given ideal gas with \(\gamma =\dfrac{{{C}_{P}}}{{{C}_{V}}}=1.5\) is at a temperature T. If the ga...

A given ideal gas with γ=CPCV=1.5\gamma =\dfrac{{{C}_{P}}}{{{C}_{V}}}=1.5 is at a temperature T. If the gas is compressed adiabatically to one-fourth of its initial volume, the final temperature will be :-
A. 2T
B. 8T
C. 22T\text{C}\text{. }2\sqrt{2}T
D. 4T

Explanation

Solution

Use the equation of state for an adiabatic process formed on a gas. You will find that the relation between the temperature and volume of the gas is constant. Then substitute the given values and find the final temperature.

Formula used:
PVa=kP{{V}^{a}}=k
PV = nRT

Complete step by step answer:
For any process performed on an ideal gas, the pressure (P) and the volume (V) of the gas are related as PVa=kP{{V}^{a}}=k, where a and k are constants. This is called an equation of state.
The value of ‘a’ depends on the process that is performed on the gas.
For an adiabatic process, a = γ\gamma .
Therefore,
⇒PVγ=k\Rightarrow P{{V}^{\gamma }}=k ….. (i).
From the ideal gas equation we know that PV = nRT,
where n is the number of moles of the gas, R is the universal gas constant and T is the temperature of the gas.
⇒P=nRTV\Rightarrow P=\dfrac{nRT}{V}.
Substitute the value of P in equation (i).
⇒(nRTV)Vγ=k\Rightarrow \left( \dfrac{nRT}{V} \right){{V}^{\gamma }}=k
⇒nRTVγ−1=k\Rightarrow nRT{{V}^{\gamma -1}}=k
⇒TVγ−1=knR\Rightarrow T{{V}^{\gamma -1}}=\dfrac{k}{nR}.
For a given gas n is constant and R is also a constant. Therefore, knR\dfrac{k}{nR} is a constant value.
This means that TVγ−1T{{V}^{\gamma -1}} is a constant value.
It is given that the initial temperature T and let its initial volume be V.
The gas is compressed to a volume equal to V4\dfrac{V}{4} and let the temperature of the gas at this time be T’.
Since TVγ−1T{{V}^{\gamma -1}} is constant,
TVγ−1=T′(V4)γ−1T{{V}^{\gamma -1}}=T'{{\left( \dfrac{V}{4} \right)}^{\gamma -1}} …. (ii).
Substitute the given value of γ\gamma =1.5 in equation (ii).
⇒TV1.5−1=T′(V4)1.5−1\Rightarrow T{{V}^{1.5-1}}=T'{{\left( \dfrac{V}{4} \right)}^{1.5-1}}
⇒TV1.5−1=T′(V1.5−141.5−1)\Rightarrow T{{V}^{1.5-1}}=T'\left( \dfrac{{{V}^{1.5-1}}}{{{4}^{1.5-1}}} \right)
⇒T=T′40.5\Rightarrow T=\dfrac{T'}{{{4}^{0.5}}}
⇒T′=40.5T=2T\Rightarrow T'={{4}^{0.5}}T=2T
This means that the final temperature of the gas is 2T.
Hence, the correct option is A.

Note:
The equation of state, i.e. PVa=kP{{V}^{a}}=k is applicable for all types of processes. However, it is applicable for an ideal gas only.
For an isothermal process, a = 1.
For an isochoric process, a = ∞\infty .
For an isobaric process, a = 0.