One mole of a monoatomic ideal gas undergoes the process A→B in the given p-V diagram. The molar heat capacity for this process is

WA→B= Area under p-V diagram =18p0V0 ΔUA→B=nCVΔT =(1)(32R)(30p0V0R−3p0V0R)=812poV0 ∴QA→B=WA→B+ΔUA→B=1172p0V0 Molar heat capacity, C=QnΔT=1172p0V0(30p0V0R)−(3p0V0R)=136R

One mole of a monoatomic ideal gas undergoes the process A→B in the given p-V diagram. The molar heat capacity for this process is

WA→B= Area under p-V diagram =18p0V0 ΔUA→B=nCVΔT =(1)(32R)(30p0V0R−3p0V0R)=812poV0 ∴QA→B=WA→B+ΔUA→B=1172p0V0 Molar heat capacity, C=QnΔT=1172p0V0(30p0V0R)−(3p0V0R)=136R

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One mole of a monoatomic ideal gas undergoes the process $A \to B$ in the given p-V diagram. The molar heat capacity for this process is

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$\frac{13 R}{6}$

$2 R$

$\frac{5 R}{2}$

$\frac{3 R}{2}$

answer is B.

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Detailed Solution

$W_{A \to B} =$ Area under p-V diagram $= 18 p_{0} V_{0}$
$Δ U_{A \to B} = n C_{V} Δ T = (1) (\frac{3}{2} R) (\frac{30 p_{0} V_{0}}{R} - \frac{3 p_{0} V_{0}}{R}) = \frac{81}{2} p_{o} V_{0} ∴ Q_{A \to B} = W_{A \to B} + Δ U_{A \to B} = \frac{117}{2} p_{0} V_{0}$
Molar heat capacity, $C = \frac{Q}{n Δ T} = \frac{\frac{117}{2} p_{0} V_{0}}{(\frac{30 p_{0} V_{0}}{R}) - (\frac{3 p_{0} V_{0}}{R})} = \frac{13}{6} R$