Question

one mole of an ideal monoatomic gas is taken through the cycle shown in the figure. the process A:B is a reversible isothermal expansion.

Given. P(A)=5atm V(A)=10litre

P(B)=1atm V(B)=50litre

P(C)=1atm V(C)=10litre

calculate the energy added to the gas.

Answer 1

Solution;

$Q_{AB}=P_AV_Aln(\frac{V_B}{V_A})$

$Q_{AB}=5×1.013×10^5×10×10^{-3}ln(\frac{50}{10})=8151.80J$

$Q_{CA}=nC_v\Delta T$

$T_A=\frac{P_AV_A}{R}=\frac{5×1.013×10^5×10×10^{-3}}{8.314}$

$T_A=609.21K$

$T_C=\frac{P_CV_C}{R}$

$T_C=\frac{1×1.031×10^5×10×10^{-3}}{8.314}$

$T_C=124.01K$

$Q_{CA}=1×\frac32×8.314(609.21-124.01)$

$Q_{CA}=6050.93J$

Total energy added;

$Q_T=Q_{AB}+Q_{CA}$

$Q_T=6050.93+8151.80$

$Q_T=14202.73J$