Question

A cylinder with a piston contains 0.150 mol of nitrogen at 1.80 * 105 Pa and 300 K. The nitrogen may be treated as an ideal gas. 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. (a) Show the series of processes in a pV-diagram. (b) Compute the temperatures at the beginning and end of the adiabatic expansion. (c) Compute the minimum p

Short answer

(a) the series of processes in a pV-diagram

(b) the temperatures at the beginning and end of the adiabatic expansion is 113.68K

(c)the minimum p is 6.84 x 10⁴Pa

Step-by-step solution

step 1:

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

Step 2:

(a) pV diagram is shown below.

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

V0 is initial volume

Step 3:

(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,

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

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.

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

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

Step 4:

(b) To find the minimum pressure,

Use,

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

Thus, the minimum p is 6.84 x 10⁴Pa