Question

A mono-atomic ideal gas (γ = 1.67, molecular weight = 40) is compressed adiabatically from 0.1 MPa, 300 K to 0.2 MPa . The universal gas constant is 8.314 kJmol^-1K^-1. The work of compression of the gas (in kJ kg^-1) is

1. 29.7
2. 19.9
3. 13.3
4. 0

Answer 1

Correct Answer - Option 1 : 29.7

Concept:

Work of compression in adiabatic process:

\(W_c=\frac{P_2V_2\;-\;P_1V_1}{1\;-\;\gamma}=\frac{P_1V_1\;-\;P_2V_2}{\gamma\;-\;1}\)

We know that PV = RT (for unit mass), therefore

\(W_c=\frac{RT_2\;-\;RT_1}{1\;-\;\gamma}=\frac{RT_1\;-\;RT_2}{\gamma\;-\;1}\)

Calculation:

Given:

P₁ = 0.1 MPa, T₁ = 300 K P₂ = 0.2 MPa

For adiabatic process:

\(\frac{{{T_2}}}{{{T_1}}} = {\left( {\frac{{{P_2}}}{{{P_1}}}} \right)^{\frac{{\gamma - 1}}{\gamma }}}\)

\(\frac{{{T_2}}}{{{300}}} = {\left( {\frac{{{0.2}}}{{{0.1}}}} \right)^{\frac{{\gamma - 1}}{\gamma }}}\)

T₂ = 396.18 K

And, \(R =\frac{\bar{R}}{M}= \frac{{8.314}}{{40}}\) = 0.207 kJ/kg-K

\(W_c= \frac{{R{T_1} - R{T_2}}}{{\gamma - 1}}\) = \(\frac{{0.207\left( {300\; -\; 396.18} \right)}}{{1\; - 1.67}}\) = -29.7 kJ/kg