Question

A perfect gas having P₁ = 0.1 N/mm², V₁ = 0.18 m³, T₁ = 20°C is compressed to \(\frac{1}{10}\) of its volume in an isothermal process. The change in entropy is

1. – 141.45 J/K
2. 141.45 J/K
3.  – 4144.23 J/K
4. Zero

Answer 1

Correct Answer - Option 1 : – 141.45 J/K

Concept:

Change in entropy for a process:

\(Δ s = {c_p} \ln \left( {\frac{{{T_2}}}{{{T_1}}}} \right) - R \ln \left( {\frac{{{P_2}}}{{{P_1}}}} \right)\)

\(Δ s = {c_v} \ln \left( {\frac{{{T_2}}}{{{T_1}}}} \right) + R\ln \left( {\frac{{{V_2}}}{{{V_1}}}} \right)\)

\(Δ s = {c_p} \ln \left( {\frac{{{V_2}}}{{{V_1}}}} \right) + {c_v}\ln \left( {\frac{{{P_2}}}{{{P_1}}}} \right)\)

where Δs = specific entropy of the process in (kJ/kg-K) and R̅ = Universal gas constant (8.314 kJ/kg-K)

As per the end-points, any-one of the given equations can be used to calculate the entropy change.

Calculation:

Given:

P1 = 0.1 N/mm2 ⇒ 0.1 MPa, V1 = 0.18 m3, \(V_2=\frac{V_1}{10}\), T1 = 20°C ⇒ 293 K

Since the process is isothermal (T₁ = T₂) and we have both the volume, we will use

\(Δ s = {c_v} \ln \left( {\frac{{{T_2}}}{{{T_1}}}} \right) + R\ln \left( {\frac{{{V_2}}}{{{V_1}}}} \right)\)

\(Δ s = R\ln \left( {\frac{{{V_2}}}{{{V_1}}}} \right)\)  .... [T₁ = T₂ = constant]

\(Δ s = 8.314 × \ln \left( {\frac{{{V_1}}}{{{10V_1}}}} \right)=-19.1437\;kJ/kgK\)

We know that PV = mR̅T

\(m=\frac{P_1V_1}{\bar RT_1}\)

\(m=\frac{0.1\;×\;10^6\;×\;0.18}{8.314 \;× \;10^3\;× \;293}=0.00739\;kg\)

ΔS = mΔs = 0.00739 × (-19.1437) = - 0.141455 kJ/K = -141.45 J/K.