91Ó°ÊÓ

piston contains of nitrogen =0.150 mol

p= 1.80 * 105 P

T=300 K

The gas is first compressed isobarically to half its original volume. It then expands adiabatically back to its original volume, and finally it is heated isochorically to its original pressure

The process is isobaric compression. The process b→c is adiabatic expansion and c→a is the isochoric heating.

V0 is initial volume

(b) the temperature at the beginning of adiabatic expansion process.

The processa→bis isobaric compression. So use,

$\frac{\mathrm{V}}{\mathrm{T}}=\frac{\mathrm{nR}}{\mathrm{p}}=\mathrm{constant}$

take the initial volume V1=V0 and the initial temperature T1=300K,

$\frac{{\mathrm{V}}_1}{{\mathrm{T}}_1}=\frac{{\mathrm{V}}_2}{{\mathrm{T}}_2}{\mathrm{pT}}_2={\mathrm{T}}_1\left(\frac{{\mathrm{V}}_2}{{\mathrm{T}}_2}\right)=300×\left(\frac{{\mathrm{V}}_0/2}{{\mathrm{V}}_0}\right)=150\mathrm{K}$

LetT3is the temperature at the end of adiabatic expansion.

Thus,

$$\begin{gathered} {\mathrm{T}}_2{{\mathrm{V}}_2}^{\mathrm{V}}−1={\mathrm{T}}_3{\mathrm{V}}_3\mathrm{Y}−1 \\ {\mathrm{V}}_2=\frac{{\mathrm{V}}_0}{2}\mathrm{and}{\mathrm{V}}_3={\mathrm{V}}_0 \end{gathered}$$

1.40γ=1.40for diatomic molecule.

${\mathrm{T}}_3={\mathrm{T}}_2{\left(\frac{{\mathrm{V}}_2}{{\mathrm{V}}_3}\right)}^{\mathrm{Y}−1}=150×{\left(\frac{{\mathrm{V}}_0/2}{{\mathrm{V}}_0}\right)}^{1.40−1)}=150×(0.51{)}^{0.4}=113.68\mathrm{K}$

Thus, the temperatures at the beginning and end of the adiabatic expansion is 113.68K

Use,

$$\begin{gathered} \frac{{\mathrm{p}}_{min}}{{\mathrm{T}}_{min}}=\frac{{\mathrm{p}}_{max}}{{\mathrm{T}}_{max}} \\ {\mathrm{T}}_{min}=114\mathrm{k},{\mathrm{p}}_{min}=1.80×{10}^5\mathrm{Pa},{\mathrm{T}}_{max}=300\mathrm{k} \\ \mathrm{So},{\mathrm{P}}_{min}={\mathrm{P}}_{max}\left(\frac{{\mathrm{T}}_{min}}{{\mathrm{T}}_{max}}\right)=1.80×{10}^5×\frac{114}{300}=6.84×{10}^4\mathrm{Pa} \end{gathered}$$

Thus, the minimum p is 6.84 x 10⁴Pa