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 expelled by the gas.

Answer 1

Solution.

$\nu=1 mol;$

$P_A=5 atm=506625Pa;$

$V_A=10L=0.01m^3;$

$P_B=1atm=101325Pa;$

$V_B=50L=0.05m^3;$

$P_C=1 atm=101325 Pa;$

$V_C=10L=0.01m^3;$

$P_AV_A=\nu RT_A\implies T_A=\dfrac{P_AV_A}{\nu R};$

$T_A=\dfrac{506625\sdot0.01}{1\sdot8.31}=609.66K;$

$P_CV_C=\nu RT_C\implies T_C=\dfrac{P_CV_C}{\nu R};$

$T_C=\dfrac{101325\sdot0.01}{1\sdot8.31}=121.93K;$

$\Delta U=\dfrac{3}{2}\nu R \Delta T;$

$\Delta U=\dfrac{3}{2}\sdot1\sdot8.31\sdot(121.93-609.66)=-6.08kJ;$

Answer: $\Delta U=-6.08kJ.$