The figure below shows a thermodynamic process followed by 0.120 g of Helium. Th
ID: 2244643 • Letter: T
Question
The figure 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 = 133 degrees C . 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 degrees 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 = 1 L and the initial temperature is T1 = 133 degrees C . Using p2 = 5p1, p3 = p1, V2 = V1, and T2 = T3 (2 --> 3 is an isothermal process) Determine the pressure p1 (in N/m), the volume V3 (in L), and the temperatures T1, and T3 (in degrees C)? How much work (W12,W23,W31) is done by the gas during each of the three segments? How much heat (Q12,Q23,Q31) is transferred to the gas during each of the three segments?Explanation / Answer
V1 = 1 L = 1*10^-3 m^3
T1 = 133+273 = 406 K
no of moles, n = 0.12/4 = 0.03 moles
P1*V1 = n*R*T1
i)
P1 = n*R*T1/V1
= 0.03*8.314*406/10^-3
= 101264.52 N/m^2
----------------------------------
P2*/T2 = P1/T1
p2 = 5*p1
T2 = 5*p1 = 5*406 = 2030 K
T3 = T2 = 2030 K
-----------------------
p2*v1 = p1*v3
v3 = (P2/P1)*v1
= 5 L
ii)
W12 = 0
W23 = n*R*T2*ln(v3/v2)
= 0.03*8.314*2030*ln(5/1)
= 815 J
W31 = -P1*(V3-V1)
= 101264.52*(5-1)*10^-3
= -405 J (here gas is compressed)
iii)
Q12 = (5/2)*n*R*(T2-T1)
= 2.5*0.03*8.314*(2030 - 406)
= 1006.4 J
Q23 = W23 = 815 J
Q31 = (5/2)*n*R*(T1-T3) + w31
= -1006.4 - 405
= -1411.4 J
= 0.03*8.314*
=
= 0.03*8.314*406/10^-3
= 101264.52 N/m^2
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