Work of 146,kJ is performed in order to compress one kilo mole of a gas adiabatically and in this pr

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11.Thermodynamics

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

A

diatomic

B

triatomic

C

a mixture of monoatomic and diatomic

D

monoatomic

Solution

$w=\frac{n R \Delta T}{\gamma-1}$

$146 \times 10^{3}=\frac{10^{3} \times 8.3 \times 7}{\gamma-1}$

$\gamma-1=\frac{8.3 \times 7}{146}=\frac{58.1}{146}$

$\gamma=\frac{58}{146}+1$

$\gamma=\frac{58+146}{146}$

$=\frac{204}{146}$

$=\frac{7}{5}$ diatomic nature.

Standard 11

Physics

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