Temperature of a monoatomic gas is increased by $△ T$ in process $P^{2} V$ = constant, Match the following two columns.

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Solution

The process is given as $P^{2} V =$ constant $⟹ P V^{1 / 2} =$ constant
Comparing with $P V^{m} =$ constant we get $m = \frac{1}{2}$

For monoatomic gas, $C_{V} = \frac{3}{2} R$
$C_{P} = C_{V} + \frac{R}{1 - m} = \frac{3}{2} R + \frac{R}{1 - \frac{1}{2}} = \frac{7}{2} R$

A ) :Work done by the gas $W = \frac{n R Δ T}{1 - m} = \frac{n R Δ T}{1 - \frac{1}{2}} = 2 n R Δ T$

B ) :Change in internal energy $Δ U = n C_{V} Δ T = \frac{3}{2} n R Δ T$

C) :Heat supplied $Q = n C_{P} Δ T = \frac{7}{2} n R Δ T$
D) : Work is done by the gas not on the gas. Hence option 4 is correct.

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