• # question_answer One mole of an ideal gas is allowed to expand reversibly and adiabatically from a temperature of $27{}^\circ C$. The work done is 3 kJ. The final temperature of the gas is equal to$[{{C}_{v}}=20k{{J}^{-1}}\,]:$ A)  450 K              B)  150 KC)  600 K                         D)  300 K

Solution :

Reversible work done during adiabatic expansion of ideal gas under the condition is $W={{\,}_{n}}{{C}_{v}}({{T}_{1}}-{{T}_{2}})$ Work done by the gas$=-3\,kJ=-3000\,J$ ${{C}_{v}}=20\,J/mol\,K$ $n=1$ ${{T}_{1}}=273+27=300\,K,{{T}_{2}}=?$ Putting the values $-3000=1\times 20({{T}_{2}}-300)$ $-150={{T}_{2}}-300$ $300-150={{T}_{2}}$ $\Rightarrow$${{T}_{2}}=150\,K$ Note: Work done during isothermal irreversible expansion of ideal gas. $W=-{{P}_{ext}}({{V}_{2}}-{{V}_{1}})$ Work done during reversible isothermal expansion of an ideal gas $W=-2.303\,nRT\log \frac{{{V}_{2}}}{{{V}_{1}}}$ Hence, the correct option is [b].

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