Verified Solution • 2025–26 Session moderate

QJEEM24Ch774f: JEE Main 2024 Chemistry - Class 11 JEEM24Ch774f Solution: An ideal gas, , is expanded adiabatically against a constant…

An ideal gas, C‾v=52R\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 K298 \mathrm{~K} and 5 atm5 \mathrm{~atm} , respectively then the final temperature is _________ K\mathrm{K} (nearest integer).

[ c‾v\overline{\mathrm{c}}_{\mathrm{v}} is the molar heat capacity at constant volume]

Step-by-Step Verified Solution

−1(2V1−V1)=n×5R2(T2−T1)-1\left(2 V_1-V_1\right)=n \times \frac{5 R}{2}\left(T_2-T_1\right)

amp;−V1=52(nRT2−5 V1)amp;−V1=2.5(nRT2)−12.5 V1amp;11.5 V1=2.5(nRT2)amp;11.5×nRT1P1=2.5×(nRT2)amp;T2=274.16 kamp;≈274 (Nearest integer) amp;\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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