question_answer

One mole of an ideal gas at an initial temperature of TK does 6R joules of work adiabatically. If the ratio of specific heats of this gas at constant pressure and at constant volume is 5/3, the final temperature of gas will be:   [AIPMT (S) 2004]

A)  \[(T+2.4)\text{ }K\]                  

B)  \[(T-2.4)\text{ }K\]

C)  \[(T+4)\text{ }K\]

D)              \[(T-4)\text{ }K\]

Correct Answer: D

Solution :

Key Idea: in an adiabatic process, there is no heat transfer into or out of a system i.e., \[Q=0\]

In an adiabatic process \[Q=0\]

So, from 1st law of thermodynamics.

\[W=-\Delta U\]

\[=-n{{C}_{V}}\Delta T\]

\[=-\left( \frac{R}{\gamma -1} \right)({{T}_{f}}-{{T}_{i}})\]

\[=\frac{nR}{\gamma -1}({{T}_{i}}-{{T}_{f}})\]

Here:     \[W=6R\,\,\,J,\,\,n=1\,\,\,mol,\]

\[R=8.31\,J/mol-K,\,\,\gamma =\frac{5}{3},{{T}_{i}}=T\,K\]

Substituting given values in Eq. (i), we get

\[\therefore \]      \[6R=\frac{R}{(5/3-1)}(T-{{T}_{f}})\]

\[\Rightarrow \]   \[6R=\frac{3R}{2}(T-{{T}_{f}})\]

\[\Rightarrow \]   \[T-{{T}_{f}}=4\]

\[\therefore \]      \[{{T}_{f}}=(T-4)K\]

Note:    Adiabatic expansions of mono, dia and polyatomic gases are shown below

1\[\to \] monoatomic

2\[\to \] diatomic

3\[\to \] polyatomic