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Question

(C) According to the first law of thermodynamics,the heat supplied $Q$ is equal to the change in internal energy $\Delta U$ plus the work done $W$: $Q

Vedclass · Accepted Answer

$\frac{2}{5}$

Vedclass · Answer

(C) According to the first law of thermodynamics,the heat supplied $Q$ is equal to the change in internal energy $\Delta U$ plus the work done $W$: $Q = \Delta U + W$.For a process at constant pressure,the heat supplied is $Q = n C_p \Delta T$.The change in internal energy is $\Delta U = n C_V \Delta T$.The work done is $W = Q - \Delta U = n C_p \Delta T - n C_V \Delta T = n R \Delta T$.The fraction of heat energy converted into work is $\frac{W}{Q} = \frac{n R \Delta T}{n C_p \Delta T} = \frac{R}{C_p}$.For a monoatomic ideal gas,$C_p = \frac{5}{2} R$.Therefore,the fraction is $\frac{W}{Q} = \frac{R}{\frac{5}{2} R} = \frac{2}{5}$.

$A$ certain amount of heat energy is supplied to a monoatomic ideal gas which expands at constant pressure. What fraction of the heat energy is converted into work?

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