The temperature of n moles of an ideal gas is increased from $T t o 4 T$ through a process for which pressure $p = a T^{- 1}$ where a is a constant . Then , the work done by the gas is ( R is universal gas constant)

n R T

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4 n R T

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2 n R T

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6 n R T

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Solution

The correct option is D $6 n R T$
Change in temperature= $Δ T = 4 T - T = 3 T$
For given process, $p T = a =$ constant
Applying ideal gas equation, $p^{2} V = c o n s t a n t \Rightarrow p V^{1 / 2} = c o n s t a n t$
So process is polytropic , i.e. of the form $p V^{x}$ =constant with $x = 1 / 2$
Now molar heat capacity=C= $\frac{R}{1 - x} + C_{v}$ , where $C_{v}$ = molar heat capacity at constant volume
$\Rightarrow C = 2 R + C_{v}$
Hence heat absorbed= $Δ Q = n C Δ T = Δ U + Δ W = n C_{v} Δ T + Δ W$
$\Rightarrow$ work done = $Δ W = n C Δ T - n C_{v} Δ T = 2 n R Δ T = 6 n R T$

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The temperature of n moles of an ideal gas is increased from T to 4T through a process for which pressure p=aT−1 where a is a constant . Then , the work done by the gas is ( R is universal gas constant)

The temperature of n moles of an ideal gas is increased from $T t o 4 T$ through a process for which pressure $p = a T^{- 1}$ where a is a constant . Then , the work done by the gas is ( R is universal gas constant)