The gure below shows a thermodynamic process followed by 0:120 g of Helium. The
ID: 2244025 • Letter: T
Question
The gure below shows a thermodynamic process followed by 0:120 g of Helium.
The initial volume is V1 = 1 L and the initial temperature is T1 = 133C . Using p2 = 5p1,
p3 = p1, V2 = V1, and T2 = T3 (2 ! 3 is an isothermal process) : (i) Determine the pressure
p1 (in N=m), the volume V3 (in L), and the temperatures T1, and T3 (in C)? (ii) How much
work (W12,W23,W31) is done by the gas during each of the three segments? (iii) How much
heat (Q12,Q23,Q31) is transferred to the gas during each of the three segments?
The figure below shows a thermodynamic process followed by 0.120 g of Helium. The initial volume is V1 = 1L and the initial temperature is T1 = 133 degree C. Using p2 = 5p1, p3 = p1, V2 = V1., and T2 = T3 (2 rightarrow 3 is an isothermal process) : (i) Determine the pressure P1 (in N/m), the volume V3 (in L), and the temperatures T1, and T3 (in degree C}? (ii) How much work (W12,W23,W31) is done by the gas during each of the three segments? (iii) How much heat (Q12,Q23,Q31) is transferred to the gas during each of the three segments?Explanation / Answer
n = 0.12/4 = 0.03
i) P1 = n*R*T1/V1 = 0.03*0.08206*(273.15+133)/1 = 1 atm = 1.013*0^5 N/m^2
P2 = 5 atm ..P3 = 1 atm.....
V3/V2 = P2/3
V3 = 5 L
T2/T1 = P2/P1 =
T2 = 5*(273.15+133) = 2030.75 K = 1757.6 oC
T3 = T2 = 2030.75 K = 1757.6 oC
ii) W12 = 0
W23 = 2.303*n*R*T2*log(V3/V2) = 2.303*0.03*8.314*5(273.15+133)*log(5)= 815.34 J
W31 = P1*(V2-V1) = 1.013*10^5*4*10^-3 = 405.2 J
iii)
Q12 = du + dW= dU = n*CV*dT = 0.03*12.47*4*(273.15+133) = 607.76286 J
Q23 = =dW = 815.34 J
Q31 = dU+dW = -n*C*dT - dW = ?1012.96286
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