Question 4.12: Problem: A gas is compressed by a piston in a cylinder from ...
Problem: A gas is compressed by a piston in a cylinder from an initial pressure of 1.0 bar and initial volume of 0.4 m³ , to a final pressure of 1.4 bar. During the compression process, the product of the gas pressure P and volume V remains constant. Determine the work done on the gas. If during the compression the gas loses 8.1 kJ of heat to the surroundings, determine the change in internal energy of the gas.
Find: Work done on the gas W and change in internal energy of the gas ∆U.
Known: Initial pressure P_1 = 1 bar, initial volume V_1 = 0.4 m³ , final pressure P_2 = 1.4 bar, heat loss Q = –8.1 kJ, during compression PV = constant.
Diagrams:
Assumptions: Changes in kinetic and potential energy are negligible, ∆KE = 0 and ∆PE = 0.
Governing Equations:
First Law Q + W = ∆U + ∆PE + ∆KE
Boundary Work W_{12}=-\int\limits_{V_1}^{V_2}{P} dV
Since ∆KE = 0 and ∆PE = 0, the energy balance reduces to
∆U = Q + W.
During the compression PV = constant = C , therefore P_1V_1 =P_2V_2 = C . The final volume is
V_2=\frac{P_1V_1}{P_2}=\frac{1.0 \ bar \times 0.4 \ m^3}{1.4 \ bar} =0.285 \ 71 \ m^3.
Work done during the compression is
W=-\int\limits_{V_1}^{V_2}{P}dV =-C\int\limits_{V_1}^{V_2}{\frac{dV}{V} } =C \ln\frac{V_1}{V_2} ,
Where the constant C can be evaluated:
C=P_1V_1=1\times 10^5 \ (N/m^2)\times 0.4 \ (m^3)=0.4 \times 10^5 \ Nm.
So, the work done is
W=C \ln\frac{V_1}{V_2} =0.4\times 10^5 \ (Nm)\times \ln\left\lgroup\frac{0.4 \ m^3}{0.285 \ 71 \ m^3} \right\rgroup =13 \ 459.5 \ J=13.460 \ kJ.
Then, the change in internal energy is
∆U = Q + W = -8.10 kJ + 13.460 kJ = 5.360 kJ.
Answer: The work done on the system during compression is 13.5 kJ. The internal energy of the system increases by 5.36 kJ.