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(C) For an ideal gas at constant pressure,the work done is given by $W = P \Delta V$.From the ideal gas equation,$PV = nRT$,so for constant pressure,$

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$\frac{2}{5} Q$

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(C) For an ideal gas at constant pressure,the work done is given by $W = P \Delta V$.From the ideal gas equation,$PV = nRT$,so for constant pressure,$P \Delta V = nR \Delta T$.Heat supplied at constant pressure is $Q = n C_p \Delta T$.For a monoatomic gas,the molar heat capacity at constant pressure is $C_p = \frac{5}{2} R$.Substituting this into the heat equation,we get $Q = n \left( \frac{5}{2} R \right) \Delta T$.Since $nR \Delta T = W$,we can write $Q = \frac{5}{2} W$.Rearranging for work done,we get $W = \frac{2}{5} Q$.

$A$ vessel contains an ideal monoatomic gas which expands at constant pressure,when heat $Q$ is given to it. Then the work done in expansion is:

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