1 mole of an ideal gas at initially (state 1) at a volume of 0.01 cubic meters a
ID: 941108 • Letter: 1
Question
1 mole of an ideal gas at initially (state 1) at a volume of 0.01 cubic meters and temperature of temperature of 300 K. Take Cv = 3/2R . Remember for an ideal gas: PV = NRT .
(a) Calculate the values of Q, W, deltaU, deltaH, deltaS, deltaA, and deltaG along the closed reversible isotherm from State 1 to State 2. State 2 is at a volume of 0.02 cubic meters.
(b) Calculate the values of Q, W, deltaU, deltaH, and deltaS along the closed isochore from State 2
to State 3. State 3 is at a temperature of 200 K.
(c) Calculate the values of Q, W, ?U, ?H, ?S, ?A, and ?G along the closed reversible isothermal compression from State 3 to State 4.
(d) Calculate the values of Q, W, ?U, ?H, and ?S along the closed isochore from State 4 back to State 1.
(e) What are the total values of Q, W, ?U, ?H, ?S, ?A, and ?G when the above cycle is run from state 1 to state 2 to state 3 to state 4 and then back to state 1?
(f) If the surrounding temperature is 300K for steps (a) and (d) and the surrounding temperature is 200K for steps (b) and (c), then what is the total entropy change of the surroundings during the cycle? Does this make sense to you?
Stirling Engine 300000 250000 200000 : 150000 100000 50000 0.0080.0 0.012 0.014 0.0160.018 0.02 0.022 Volume (m3)Explanation / Answer
1 mole of an ideal gas at initially (state 1) at a volume of 0.01 cubic meters and temperature of temperature of 300 K. Take Cv = 3/2R . Remember for an ideal gas: PV = NRT .
(a) Calculate the values of Q, W, deltaU, deltaH, deltaS, deltaA, and deltaG along the closed reversible isotherm from State 1 to State 2. State 2 is at a volume of 0.02 cubic meters.
Answer: The step is expanstion of gas from 0.01 to 0.02
So the work will be done by the system which will be
W =- nRT ln(V2/V2) = 1X0.8.314 X 300 ln(2) = -1728.48 Joules
Internal energy change will be zero(isothermal) and the heat will be equal to work
Q = -W = 1728.48 J
DeltaS = 1728.48 J / T = 5.76 J
Delta H = 0
DeltaG = -TdeltaS = 300 X 1728 =518KJ
(b) Calculate the values of Q, W, deltaU, deltaH, and deltaS along the closed isochore from State 2
to State 3. State 3 is at a temperature of 200 K.
Answer: Isochoric means constant volume
Work done = 0
Heat = Internal energy = Cv (T2-T1) = 3/2R (300-100) = 8.314 x 1.5 X 100 =1 247.1 Joules
Enthalpy = qv + V delta P
so if we know the change in volume we can calcualte enthalpy (not clear from diagram)
Entropy = Cv ln T2/T1 = 1247.1 X ln (2/3) = 12.039 KJ
(c) Calculate the values of Q, W, U, H, S, A, and G along the closed reversible isothermal compression from State 3 to State 4.
Answer: The step is compression of gas from 0.02 to 0.01
So the work will be done by the system which will be
W = nRT ln(V2/V2) = 1X0.8.314 X 300 ln(2) = 1728.48 Joules
Internal energy change will be zero(isothermal) and the heat will be equal to work
Q = W = 1728.48 J
DeltaS = 1728.48 J / T = 5.76 J
Delta H = 0
DeltaG = 1728.48 J
(d) Calculate the values of Q, W, U, H, and S along the closed isochore from State 4 back to State 1.
Answer: Isochoric means constant volume
Work done = 0
Heat = Internal energy = -Cv (T2-T1) = -3/2R (300-200) = -8.314 x 1.5 X 100 =-1 247.1 Joules
Enthalpy = qv + V delta P
so if we know the change in volume we can calcualte enthalpy (not clear from diagram)
Entropy = Cv ln T2/T1 = 1247.1 X ln (2/3) = -12.039 KJ
(e) What are the total values of Q, W, U, H, S, A, and G when the above cycle is run from state 1 to state 2 to state 3 to state 4 and then back to state 1?
Change in internal energy = 0
Heat = sum of two heat
Work = 0
(f) If the surrounding temperature is 300K for steps (a) and (d) and the surrounding temperature is 200K for steps (b) and (c), then what is the total entropy change of the surroundings during the cycle? Does this make sense to you?
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