MHT CET · Physics · Thermodynamics
One mole of an ideal gas at an initial temperature of ' \(T\) ' K does ' 6 R ' 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 (\(\left.\mathrm{R}=8.31 \mathrm{~J} \mathrm{~mole}^{-1} \mathrm{~K}^{-1}\right)\)
- A \((T+4 \cdot 2) K\)
- B \((\mathrm{T}-4 \cdot 2) \mathrm{K}\)
- C \(\quad(T+4) K\)
- D \((T-4) K\)
Answer & Solution
Correct Answer
(D) \((T-4) K\)
Step-by-step Solution
Detailed explanation
For an adiabatic process,
\(\mathrm{W}=\frac{\mathrm{nR}\left(\mathrm{~T}_{\mathrm{i}}-\mathrm{T}_{\mathrm{f}}\right)}{\gamma-1}\)
\(\begin{array}{ll}\therefore \quad & 6 R=\frac{R\left(T-T_f\right)}{\left(\frac{5}{3}-1\right)} \quad \ldots .\text{(Given: } \mathrm{n}=1) \\ \therefore & T_f=(T-4) K\end{array}\)
\(\mathrm{W}=\frac{\mathrm{nR}\left(\mathrm{~T}_{\mathrm{i}}-\mathrm{T}_{\mathrm{f}}\right)}{\gamma-1}\)
\(\begin{array}{ll}\therefore \quad & 6 R=\frac{R\left(T-T_f\right)}{\left(\frac{5}{3}-1\right)} \quad \ldots .\text{(Given: } \mathrm{n}=1) \\ \therefore & T_f=(T-4) K\end{array}\)
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