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 $\mathrm{C_V}=20 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$
Answer (integer)
150
Solution
<p>$\mathrm{T_1=300~K}$</p>
<p>$\mathrm{w = 3}$ kJ/mole</p>
<p>$\mathrm{w=nC_v\Delta T}$</p>
<p>$3000=1\times20\times(300-\mathrm{T_2})$</p>
<p>$300-\mathrm{T_2}=150$</p>
<p>$\mathrm{T_2=150~K}$</p>
About this question
Subject: Chemistry · Chapter: Thermodynamics · Topic: Zeroth and First Law
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