JEE MAIN - Chemistry (2024 - 6th April Morning Shift - No. 24)
An ideal gas, $$\overline{\mathrm{C}}_{\mathrm{v}}=\frac{5}{2} \mathrm{R}$$, is expanded adiabatically against a constant pressure of 1 atm untill it doubles in volume. If the initial temperature and pressure is $$298 \mathrm{~K}$$ and $$5 \mathrm{~atm}$$, respectively then the final temperature is _________ $$\mathrm{K}$$ (nearest integer).
[$$\overline{\mathrm{c}}_{\mathrm{v}}$$ is the molar heat capacity at constant volume]
Explanation
$$-1\left(2 V_1-V_1\right)=n \times \frac{5 R}{2}\left(T_2-T_1\right)$$
$$\begin{aligned} & -\mathrm{V}_1=\frac{5}{2}\left(n R T_2-5 \mathrm{~V}_1\right) \\ & -\mathrm{V}_1=2.5\left(\mathrm{nRT_{2 } )}-12.5 \mathrm{~V}_1\right. \\ & 11.5 \mathrm{~V}_1=2.5\left(\mathrm{nRT_{2 } )}\right. \\ & 11.5 \times \frac{n R T_1}{P_1}=2.5 \times\left(n R T_2\right) \\ & \mathrm{T}_2=274.16 \mathrm{~k} \\ & \approx 274 \text { (Nearest integer) } \\ & \end{aligned}$$
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