JEE MAIN - Chemistry (2023 - 30th January Evening Shift - No. 19)
1 mole of ideal gas is allowed to expand reversibly and adiabatically from a temperature of $27^{\circ} \mathrm{C}$. The work done is $3 \mathrm{~kJ} \mathrm{~mol}^{-1}$. The final temperature of the gas is ________ $\mathrm{K}$ (Nearest integer).
Given $\mathrm{C_V}=20 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$
Given $\mathrm{C_V}=20 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$
Answer
150
Explanation
$$\mathrm{T_1=300~K}$$
$$\mathrm{w = 3}$$ kJ/mole
$$\mathrm{w=nC_v\Delta T}$$
$$3000=1\times20\times(300-\mathrm{T_2})$$
$$300-\mathrm{T_2}=150$$
$$\mathrm{T_2=150~K}$$
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