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:$\frac{\Delta Q}{\Delta W} = \frac{1}{1 - \frac{3}{5}} = \frac{1}{\frac{2}{5}} = \frac{5}{2} = 2.5$.

Column-$I$	Column-$II$
$(a)$ Horizontal line parallel to $V$-axis	$(i)$ Adiabatic process
$(b)$ Vertical line parallel to $P$-axis	$(ii)$ Isobaric process
	$(iii)$ Isochoric process

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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