A gas is contained in a piston-cylinder assembly. The gas undergoes a polytropic

Question

A gas is contained in a piston-cylinder assembly. The gas undergoes a polytropic process from the state 1(p1=2 bar amd V1 unknown) to the state 2 (p2=8 bar and V2= 0.02m^3), and the relation between pressure and volume is p.V^1.3= constant (n=1.3). The mass of the gas is 0.2kg, and the specific internal energy of the gas increases by 50kJ per kg during the process. Consider the gas is the thermodynamic system, and the changes in kinetic energy and potential energy are negligible, please determine:

a) The initial colume of the gas, V1, in m^3

b)The work during the process, W, in kJ;

c)Please identify the direction of energy transfer by work: into gas or out of gas

d) The change in internal energy of the gas, U, in kJ

e) The heat transfer during the process, Q, in kJ

f) Please judge the process for the gas: exothermic or endothermic

Explanation / Answer

P1=2 bar
P2=8 bar
V2=0.02 m3
n = 1.3
Sign convention:
W is +ve work done on the system by compression
Q is +ve heat is given to the system

a) The initial colume of the gas, V1, in m^3
P1*V1^1.3 = P2*V2^1.3
V1 = V2*(P2/P1)^(1/1.3)=0.02*(8/2)^0.77=0.058 m3

b)The work during the process, W, in kJ;
W = -(P1*V1-P2*V2)/(n-1) = -(2*0.058-8*0.02)/(1.3-1) = 0.15 bar*m3
1bar = 1E5 N/m2
W = 0.15*1E5 = 15E3 J = 15 kJ

c)Please identify the direction of energy transfer by work: into gas or out of gas
W is positive. So work is done into gas due to compression of volume from 0.058 m3 to 0.02 m3.

d) The change in internal energy of the gas, U, in kJ
Given change in specific internal energy, du = 50 kJ/kg
Mass of gas, w = 0.2 kg
Change in total internal energy, dU = w*du = 0.2*50 = 10 kJ

e) The heat transfer during the process, Q, in kJ
Q = dU - W = 10 - 15 = -5 kJ

f) Please judge the process for the gas: exothermic or endothermic
As Q is negative, heat comes out of the system. So it is exothermic.

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