Two moles of nitrogen are initially at 10 bar and 600 K (state 1) in a horizonta
ID: 2994056 • Letter: T
Question
Two moles of nitrogen are initially at 10 bar and 600 K (state 1) in a horizontal piston/cylinder device. They are expanded adiabatically to 1 bar (state 2). They are then heated at constant volume to 600 K (state 3). Finally, they are isothermally returned to state 1. Assume that N2 is an ideal gas with a constant heat capacity as given on the back flap of the book. Neglect the heat capacity of the piston/cylinder device. Suppose that heat can be supplied or rejected as illustrated below. Assume each step of the process is reversible.
Calculate the heat transfer and work done on the gas for each step and overall.
The atmosphere is at 1 bar and 298 K throughout the process. Calculate the work done on the atmosphere for each step and overall. (Hint: Take the atmosphere as the system.) How much work is transferred to the shaft in each step and overall?
Two moles of nitrogen are initially at 10 bar and 600 K (state 1) in a horizontal piston/cylinder device. They are expanded adiabatically to 1 bar (state 2). They are then heated at constant volume to 600 K (state 3). Finally, they are isothermally returned to state 1. Assume that N2 is an ideal gas with a constant heat capacity as given on the back flap of the book. Neglect the heat capacity of the piston/cylinder device. Suppose that heat can be supplied or rejected as illustrated below. Assume each step of the process is reversible. Calculate the heat transfer and work done on the gas for each step and overall. Taking state 1 as the reference state, and setting , calculate U and H for the nitrogen at each state, and ?U and ?H for each step and the overall Q and WEC. The atmosphere is at 1 bar and 298 K throughout the process. Calculate the work done on the atmosphere for each step and overall. (Hint: Take the atmosphere as the system.) How much work is transferred to the shaft in each step and overall?Explanation / Answer
T2 / T1 = (P2 / P1)^((n-1)/n)
T2 / 600 = (1 / 10)^((1.4-1)/1.4)
T2 = 310.8 K
P2 / T2 = P3 / T3
1 / 310.8 = P3 / 600
P3 = 1.93 bar
Heat transfer Q12 = 0 (since adiabatic)
Work done W12 = nR(T2 - T1) / (1-n)
= 2*8.314*(310.8 - 600) / (1-1.4)
= 12022 J
= 12.022 kJ
Q23 = n*Cv*(T3 - T2) = nR(T3 - T2) / (n-1) = 2*8.314*(600 - 310.8) / (1.4 - 1) = 12022 J = 12.022 kJ
W23 = 0 (since constant volume)
Q31 = nRT3 ln (P1 / P3) = 2*8.314*600 ln(10 / 1.93) = 16412.5 J = 16.412 kJ
W31 = Q31 = 16.412 kJ
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