Q134 / 150 · JEE Main 2006 · Physics

The work of $146 \mathrm{~kJ}$ is performed in order to compress one kilo mole of gas adiabatically and in this process the temperature of the gas increases by $7^{\circ} \mathrm{C}$. The gas is $\left(\mathrm{R}=8.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right)$

  • A
    monoatomic
  • Correct answer
    B
    diatomic
  • C
    triatomic
  • D
    a mixture of monoatomic and diatomic

Solution

$146=C_v \Delta T$ $\Rightarrow C_v=21 \mathrm{~J} / \mathrm{mol} \mathrm{K}$

Hence, the correct answer is option: 'B'.

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