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(B) For an ideal gas,the equation of state is $PV = nRT$. Since the pressure $P$ is constant,the change in volume $\Delta V$ due to a change in temper

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$40$

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(B) For an ideal gas,the equation of state is $PV = nRT$. Since the pressure $P$ is constant,the change in volume $\Delta V$ due to a change in temperature $\Delta T$ is given by $P \Delta V = nR \Delta T$. The work done $W$ in a process at constant pressure is defined as $W = P \Delta V$. Substituting the expression for $P \Delta V$,we get $W = nR \Delta T$. Given $n = 2$ moles and $\Delta T = 20 \ K$ (or $20^{\circ} C$),the work done is $W = 2 	imes R 	imes 20 = 40R$.

When the temperature of $2$ moles of an ideal gas is increased by $20^{\circ} C$ at constant pressure,find the work involved in the process. (in $R$)

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