Physics - Heat & Thermo Question with Solution | TestHub

For process $\mathrm{AB}{\mathrm{T}}_{\mathrm{A}}=300\mathrm{K},{\mathrm{T}}_{\mathrm{B}}=600\mathrm{K}$

$\mathrm{W}=\mathrm{nR}\Delta \mathrm{T}=\mathrm{nR}\left({\mathrm{T}}_{\mathrm{B}}-{\mathrm{T}}_{\mathrm{A}}\right)=300\mathrm{nR}=600\mathrm{R}$.

$Q=nCp\Delta T=2\times \frac{5}{2}R(300)=1500R.$

For process BC

$\begin{array}{rlr}& W=nRT\ln \frac{{\mathrm{V}}_{\mathrm{f}}}{{\mathrm{V}}_{\mathrm{c}}}=nRT\ln \frac{{\mathrm{p}}_{\mathrm{i}}}{{\mathrm{p}}_{\mathrm{f}}}=nRT\ln 2=1200R\ln 2& \end{array}$

$Q=W=1200R\ln 2$

For process CA

$$\begin{array}{rlrlrlrl}W=& \int PdV={\int }_{600}^{300}\frac{\mathrm{K}}{\mathrm{T}}\frac{2\mathrm{nRT}}{\mathrm{K}}dT.& & & & & & \end{array}$$

$$=-2nR(300)=-1200R.$$

$$Q=n{C}_V\Delta T+W=2\times \frac{3}{2}R(-300)-1200R.$$

$$=-900R-1200R=-2100R$$

$$n=\frac{600R+1200R\ln 2-1200R}{1500R+1200R\ln 2}=1-\frac{21}{12\ln 2+15}\Rightarrow x=7.$$