When an ideal gas left(gamma = frac{5}{3}righ | vedclass

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(D) For an ideal gas undergoing a process at constant pressure,the heat supplied $dQ$ is given by $dQ = n C_p dT$.The work done by the gas is $dW = P

Vedclass · Accepted Answer

$40$

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(D) For an ideal gas undergoing a process at constant pressure,the heat supplied $dQ$ is given by $dQ = n C_p dT$.The work done by the gas is $dW = P dV = n R dT$.The fraction of heat energy utilized in doing external work is $\frac{dW}{dQ} = \frac{n R dT}{n C_p dT} = \frac{R}{C_p}$.We know that $C_p = \frac{\gamma R}{\gamma - 1}$.Substituting this,we get $\frac{dW}{dQ} = \frac{R}{\frac{\gamma R}{\gamma - 1}} = \frac{\gamma - 1}{\gamma} = 1 - \frac{1}{\gamma}$.Given $\gamma = \frac{5}{3}$,we have $\frac{dW}{dQ} = 1 - \frac{1}{5/3} = 1 - \frac{3}{5} = \frac{2}{5}$.Converting to percentage: $\frac{2}{5} 	imes 100 \% = 40 \%$.

When an ideal gas $\left(\gamma = \frac{5}{3}\right)$ is heated under constant pressure,then what percentage of given heat energy will be utilised in doing external work (in $\%$)?

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