146=CvΔT ⇒Cv=21 J/molK
JEE Main 2006 — Physics Thermodynamics
The work of 146 kJ is performed in order to compress one kilo mole of gas adiabatically and in this process the temperature of the gas increases by 7∘C. The gas is (R=8.3 J mol−1 K−1)
Held on 30 Apr 2006 · Verified 6 Jul 2026.
monoatomic
diatomic
triatomic
a mixture of monoatomic and diatomic
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10 mole of an ideal gas is undergoing the process shown in the figure. The heat involved in the process from $P_{1}$ to $P_{2}$ is $\alpha$ Joule ($P_{1}=21.7 \mathrm{~Pa}$ and $\left.P_{2}=30 \mathrm{~Pa}, \mathrm{C}_{v}=21 \mathrm{~J} / \mathrm{K}. \mathrm{mol}, R=8.3 \mathrm{~J} / \mathrm{mol}. \mathrm{K}\right)$. The value of $\alpha$ is $\_\_\_\_$. 
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