An ideal gas is heated at constant pressure. | vedclass

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(A) For an ideal gas heated at constant pressure,the heat absorbed is given by $Q = n C_P \Delta T$.The change in internal energy is given by $\Delta

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$1 - \frac{1}{\gamma }$

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(A) For an ideal gas heated at constant pressure,the heat absorbed is given by $Q = n C_P \Delta T$.The change in internal energy is given by $\Delta U = n C_V \Delta T$.According to the first law of thermodynamics,$Q = \Delta U + W$,so the work done is $W = Q - \Delta U$.The ratio of work done $(W)$ to the heat absorbed $(Q)$ is:$\frac{W}{Q} = \frac{Q - \Delta U}{Q} = 1 - \frac{\Delta U}{Q}$.Substituting the expressions for $Q$ and $\Delta U$:$\frac{W}{Q} = 1 - \frac{n C_V \Delta T}{n C_P \Delta T} = 1 - \frac{C_V}{C_P}$.Since the adiabatic index is defined as $\gamma = \frac{C_P}{C_V}$,we have $\frac{C_V}{C_P} = \frac{1}{\gamma}$.Therefore,the ratio is $1 - \frac{1}{\gamma}$.

An ideal gas is heated at constant pressure. If $Q$ is the amount of heat absorbed,then the ratio of the work done to the heat absorbed for increasing the internal energy is:

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