8.) One mole of an ideal monatomic gas (C, m 3/2)is through a cycle consisting o
ID: 705466 • Letter: 8
Question
8.) One mole of an ideal monatomic gas (C, m 3/2)is through a cycle consisting of the following three reversible steps: (A) Isothermal compression from 2 atm and 10 L to 20 atm and 1 L. (B) Isobaric expansion to return the gas to the original volume of 10 L with T going n/from T? to T2 h.1 (C) cooling at constant volume to bring the gas to the original pressure and temperature. Illustrate the steps on a P vs. V curve. Set V as your abcissa. Calculate T1 and T2. Calculate ?, ?U, q, and w in joules for each step and for the cycle.Explanation / Answer
8) We have one mole of the ideal gas with Cv,m = 3R/2.
We know that, for an ideal gas,
Cp,m – Cv,m = R
====> Cp,m = Cv,m + R = 3R/2 + R = 5R/2
A) Determine the initial temperature T1 of the gas first. We have,
P = nRT/V
====> T1 = P1V1/nR = (2 atm)*(10 L)/(1 mole)*(0.082 L-atm/mol.K) = 243.90 K.
The gas undergoes an isothermal compression in volume from 2 atm and 10 L to 20 atm and 1 L.
Since the process is isothermal, hence, ?T (change in temperature) = 0.
For an ideal gas, the change in internal energy, ?U depends only on the temperature. Since ?T = 0, we must have ?U = 0 (ans)
For an ideal gas, we have
P1*V1 = P2*V2
where the P and V terms denote the pressure and volume of the gas at the initial condition (denote as i) and the final condition (denote as f).
Therefore,
V2/V1 = P1/P2……(1)
The change in enthalpy for an ideal gas is given as
dH = dU + PdV + VdP
For an ideal gas, we have
P = nRT/V
and V = nRT/P
Therefore,
?dH = ?dU + nRT?dV/V + nRT?dP/P
====> ?H = ?U + nRT*ln V2/V1 + nRT*ln P2/P1
====> ?H = 0 + nRT*ln P1/P2 + nRT*ln P2/P1 [?U = 0]
====> ?H = nRT*ln P1/P2 + nRT*ln 1/(P1/P2)
====> ?H = nRT*ln P1/P2 - nRT*ln P1/P2 = 0 (ans)
The work done by the gas is given as
dw = -PdV
Again, P = nRT/V
Therefore, dw = -nRT*dV/V
Integrating,
?dw = -nRT*?dV/V
====> w = -nRT1*ln V2/V1 = -(1 mole)*(0.082 L-atm/mol.K)*(243.90 K)*ln (1 L/10 L)
= -(19.9998 L-atm)*(-2.303) = 46.059 L-atm
= (46.059 L-atm)*(101.33 J/1 L-atm) = 4667.1585 J
= (4667.1585 J)*(1 kJ/1000 J) = 4.6671585 kJ ? 4.7 kJ (1 L-atm = 101.33 J) (ans).
As per the first law, we have,
?U = q + w
Since ?U = 0, we have,
q + w = 0
====> q = -w = -4.7 kJ (ans).
B) Determine T2. We have the pressure P constant (isobaric = constant pressure). As per Charle’s law,
V1/T1 = V2/T2 (here V1 = 1 L and V2 = 10 L)
====> T2 = V2*T1/V1 = (10 L)*(243.90 K)/(1 L) = 2439.00 K.
We have the change in internal energy
?U = n*Cv,m*(T2 – T1) = (1 mole)*(3R/2)*(2439.00 – 243.90) K
====> ?U = (1 mole)*(3/2*0.082 L-atm/mol.K)*(2195.10 K) = 269.9973 L-atm
= (269.9973 L-atm)*(101.33 J/1 L-atm) = 27358.8264 J
= (27358.8264 J)*(1 kJ/1000 J) = 27.3588264 kJ ? 27.4 kJ (ans).
The enthalpy change is again given as
?H = ?U + P?V + V?P = ?U + P?V (?P = 0, isobaric process)
= ?U + nR?T (ideal gas law)
= 27.4 kJ + (1 mole)*(0.082 L-atm/mol.K)*(2439.00 – 243.90) K
= 27.4 kJ + 179.9982 L-atm
= 27.4 kJ + (179.9982 L-atm)*(101.33 J/1 L-atm)
= 27.4 kJ + 18239.21761 J
= 27.4 kJ + (18239.21761 J)*(1 kJ/1000 J)
= 27.4 kJ + 18.23921761 kJ
? 27.4 kJ + 18.2 kJ = 45.6 kJ (ans).
The work done in the isobaric process is
w = -P?V = -nR?T
= -(1 mole)*(0.082 L-atm/mol.K)*(2439.00 – 243.90) K
= -179.9982 L-atm
= -(179.9982 L-atm)*(101. 33 J/1 L-atm) = -18293.21761 J
= -(18239.21761 J)*(1 kJ/1000 J)
= -18.23921761 kJ ? -18.2 kJ (ans).
As per the first law,
?U = q + w
====> q = ?U – w = (27.4 kJ) – (-18.2 kJ) = 45.6 kJ (ans).
C) The volume is kept constant; hence, ?V = 0. Also, the final temperature is T2 = 243.90 K and the final volume is P2 = 2 atm. Since ?V = 0, we have, w = -P?V = 0 (ans).
The internal energy change depends only on the change in temperature; therefore,
?U = n*Cv,m*(T2 – T1)
= (1 mole)*(3R/2)*(243.90 – 239.00) K
= (1 mole)*(3/2*0.082 L-atm/mol.K)*(-2195.10 K) = -269.9973 L-atm
= -(269.9973 L-atm)*(101.33 J/1 L-atm) = -27358.8264 J
= -(27358.8264 J)*(1 kJ/1000 J) = -27.3588264 kJ ? -27.4 kJ (ans).
The change in enthalpy is given as
?H = ?U + P?V + V?P = ?U + V?P (?V = 0)
= ?U + nR?T
= ?U + (1 mole)*(0.082 L-atm/mol.K)*(243.90 – 2439.00) K
= ?U + (-179.9982 L-atm)
= -27.4 kJ - (179.9982 L-atm)*(101.33 J/1 L-atm)
= -27.4 kJ - 18239.21761 J
= -27.4 kJ - (18239.21761 J)*(1 kJ/1000 J)
= -27.4 kJ - 18.23921761 kJ
? -27.4 kJ - 18.2 kJ = -45.6 kJ (ans).
As per the first law, we have
?U = q + w
====> q = ?U = -27.4 kJ (ans).
The overall change in internal energy is given as
?Utot = 0 + (27.4 kJ) + (-27.4 kJ) = 0 (ans)
The overall change in enthalpy is given as
?Htot = 0 + (45.6 kJ) + (-45.6 kJ) = 0 (ans).
The overall work is given as
wtot = (4.7 kJ) + (-18.2 kJ) = -13.5 kJ (ans).
The overall heat involved is given as
qtot = (-4.7 kJ) + (45.6 kJ) + (-27.4 kJ) = +13.5 kJ (ans).
Plot the data. The volume is on the absicca (x-axis) and the pressure is on the ordinate (y-axis).
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