Volume versus temperature graph of two moles | vedclass

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(B) The $V-T$ graph is a straight line passing through the origin. According to the ideal gas law $PV = nRT$,if $V \propto T$,then the pressure $P$ re

Vedclass · Accepted Answer

$2.5$

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(B) The $V-T$ graph is a straight line passing through the origin. According to the ideal gas law $PV = nRT$,if $V \propto T$,then the pressure $P$ remains constant. Thus,the process $1-2$ is an isobaric process.For an isobaric process,the heat absorbed is given by $\Delta Q = nC_P \Delta T$.The work done by the gas is given by $\Delta W = P \Delta V = nR \Delta T$.Using the relation $C_P - C_V = R$,we can write $\Delta W = n(C_P - C_V) \Delta T$.The ratio of heat absorbed to the work done is:$\frac{\Delta Q}{\Delta W} = \frac{nC_P \Delta T}{n(C_P - C_V) \Delta T} = \frac{C_P}{C_P - C_V} = \frac{1}{1 - \frac{C_V}{C_P}} = \frac{1}{1 - \frac{1}{\gamma}}$For a monoatomic gas like helium,the adiabatic index $\gamma = \frac{C_P}{C_V} = \frac{5}{3}$.Therefore,the ratio is:$\frac{\Delta Q}{\Delta W} = \frac{1}{1 - \frac{3}{5}} = \frac{1}{\frac{2}{5}} = \frac{5}{2} = 2.5$.

Volume versus temperature graph of two moles of helium gas is as shown in figure. The ratio of heat absorbed and the work done by the gas in process $1-2$ is

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