One mole of an ideal gas, initially at 30degreeC and 1 bar, is changed to 130deg
ID: 977142 • Letter: O
Question
One mole of an ideal gas, initially at 30degreeC and 1 bar, is changed to 130degreeC and 10 bar by three different mechanically reversible processes: The gas is first heated at constant volume until its temperature is 130degreeC; then it is compressed isothermally until its pressure is 10 bar. The gas is first heated at constant pressure until its temperature is 130degreeC; then it is compressed isothermally to 10 bar. The gas is first compressed isothermally to 10 bar; then it is heated at constant pressure to 130degreeC. Calculate Q,W, DeltaU, and, DeltaH in each case Take C_p (7/2) R C_v= (5/2)R Alternatively, take C_p = (5/2)R andC_v =(3/2)R.Explanation / Answer
1. i) heating at constant volume:
As dV = 0, dW = -PdV = 0
dU = dq + dW = dq
dU = dq
= n Cv dT
= 1 mol * 3/2 R * (403-303)K
= 1.2741 kJ (put R = 8.314 J/mol K)
dH = dU + dPV = dU + nRdT
= 1.2741 kJ + [ 1 mol * 8.314 J/mol K * 100 K]
= 2.0785 kJ
ii) isothermal compression:
As the process is isothermal, dU= dH = 0
So, dq = - dW = - pdV
V2 = nRT/P2 = (1 mol * 0.082 lit.atm/mol K * 403 K) / 9.87 atm [as 10 bar = 9.87 atm]
= 3.35 lit
V1 = nRT/P1 = (1 mol * 0.082 lit.atm/mol K * 403 K) / 0.987 atm [as 1 bar = 0.987 atm]
= 33.5 lit
dW = - pdV = - 0.987 atm (3.35 - 33.5)lit
= 29.78 lit.atm
= 3.015 kJ
q = - 3.015 kJ
Total:
dU = 1.2741 kJ
dH = 2.0785 kJ
Q = 1.2741 kJ + (- 3.015 kJ)
= -1.741 kJ
W = 3.015 kJ
2. i) heating at constant pressure:
dH = nCpdT = 1 mol * 5/2 R * 100 K
= 2.0785 kJ
W = -dPV = - nRdT
= - 1 mol * 8.314 J/mol K * 100 K
= - 0.8314 kJ
dU = dH - dPV = 2.0785 kJ - 0.8314 kJ = 1.2471 kJ
q = dU - W = 1.2471 kJ - (- 0.8314 kJ) = 2.0785 kJ
ii) isothermal compression:
As the process is isothermal, dU= dH = 0
So, dq = - dW = - pdV
So, dq = - dW = - pdV
V2 = nRT/P2 = (1 mol * 0.082 lit.atm/mol K * 403 K) / 9.87 atm [as 10 bar = 9.87 atm]
= 3.35 lit
V1 = nRT/P1 = (1 mol * 0.082 lit.atm/mol K * 403 K) / 0.987 atm [as 1 bar = 0.987 atm]
= 33.5 lit
dW = - pdV = - 0.987 atm (3.35 - 33.5)lit
= 29.78 lit.atm
= 3.015 kJ
q = - 3.015 kJ
Total:
dU = 1.2471 kJ
dH = 2.0785 kJ
Q = 2.0785 kJ + (- 3.015 kJ) = -0.9365 kJ
W = (- 0.8314 kJ) + 3.015 kJ
= 2.1836 kJ
3. i) isothermal compression:
As the process is isothermal, dU= dH = 0
So, dq = - dW = - pdV
V2 = nRT/P2 = (1 mol * 0.082 lit.atm/mol K * 303 K) / 9.87 atm [as 10 bar = 9.87 atm]
= 2.517 lit
V1 = nRT/P1 = (1 mol * 0.082 lit.atm/mol K * 303 K) / 0.987 atm [as 1 bar = 0.987 atm]
= 25.173 lit
dW = - pdV = - 0.987 atm (2.517 - 25.173)lit
= 22.3617 lit.atm
= 2.2658 kJ
q = - 2.2658 kJ
ii) Heating at constant pressure:
dH = nCpdT = 1 mol * 5/2 R * 100 K
= 2.0785 kJ
W = -dPV = - nRdT
= - 1 mol * 8.314 J/mol K * 100 K
= - 0.8314 kJ
dU = dH - dPV = 2.0785 kJ - 0.8314 kJ = 1.2471 kJ
q = dU - W = 1.2471 kJ - (- 0.8314 kJ) = 2.0785 kJ
Total:
dU = 1.2471 kJ
dH = 2.0785 kJ
Q = (- 2.2658 kJ) + 2.0785 kJ
= -0.1873 kJ
W = 2.2658 kJ + (- 0.8314 kJ)
= 1.4344 kJ
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