Q#9.18
During an isothermal compression of an ideal gas, 335 J of heat must be removed from the gas to maintain constant temperature. How much work is done by the gas during the process?
Answer:
We can use ΔU = Q − W
Q < 0 when heat leaves the gas.
For an isothermal process, ΔU = 0, so W = Q = - 335 J
In a compression the volume decreases and W < 0.
Q#19.19 .
A cylinder contains 0.250 mol of carbon dioxide (CO$_2$) gas at a temperature of 27.0 $^0$ C. The cylinder is provided with a frictionless piston, which maintains a constant pressure of 1.00 atm on the gas. The gas is heated until its temperature increases to 127.0 $^0$ C. Assume that the $CO_2$ may be treated as an ideal gas.
(a) Draw a pV-diagram for this process.
(b) How much work is done by the gas in this process?
(c) On what is this work done?
(d) What is the change in internal energy of the gas?
(e) How much heat was supplied to the gas?
(f) How much work would have been done if the pressure had been 0.50 atm?
Answer:
For a constant pressure process,
W = pΔV, Q = n$C_v\Delta T$ and $\Delta U = nC_v \Delta T$
ΔU = Q − W and $C_p=C_v+R$. For an ideal gas, pΔV = nRΔT
Given: For carbon dioxide (CO$_2$), $C_v$ = 28.46 J/mol.K
(a) The pV diagram is given in Figure 19.19.
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| Fig.19.9 |
W = 208 J
(c) The work is done on the piston.
(d) Since Eq. $\Delta U = nC_v \Delta T$ holds for any process,
$\Delta U = (0.250 mol)(28.48 J/mol.K)(127.0 $^0$C - 27.0 $^0$ C) = 712 J
(e) Either W = n$C_p\Delta T$ or Q = ΔU + W gives
Q = 208 J + 712 J = 920 J to three significant figures.
(f) The lower pressure would mean a correspondingly larger volume, and the net result would be that the work done would be the same as that found in part (b).
W = nRΔT, so W, Q and ΔU all depend only on ΔT. When T increases at constant pressure, V increases and W > 0. ΔU and Q are also positive when T increases.
Q#19.20
A cylinder contains 0.0100 mol of helium at T = 27.0$^0C$
(a) How much heat is needed to raise the temperature to 67.0 $^0$C while keeping the volume constant? Draw a pV-diagram for this process.
(b) If instead the pressure of the helium is kept constant, how much heat is needed to raise the temperature from 27.0 $^0$ C to 67.0 $^0$C. Draw a pV-diagram for this process.
(c) What accounts for the difference between your answers to parts (a) and (b)? In which case is more heat required? What becomes of the additional heat?
(d) If the gas is ideal, what is the change in its internal energy in part (a)? In part (b)? How do the two answers compare? Why?
Answer:
For constant volume Q = n$C_v\Delta T$
For constant pressure Q = n$C_p\Delta T$
For any process of an ideal gas $\Delta U = nC_v \Delta T$
Given: For helium, $C_v$ = 12.47 J/mol.K and $C_p$ = 20.78 J/mol.K
(a) Q = n$C_v\Delta T$
Q = (0.0100 mol)(12.47 J/mol.K)(67.0 $^0$C - 27.0 $^0$ C) = 4.99 J
The pV-diagram is sketched in Figure 19.20a.
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| Fig.20 |
(b) Q = n$C_p\Delta T$ =
Q = (0 0100 mol)(20.78 J/mol K)(40.0$^0$C) = 8.31 J.
The pV-diagram is sketched in Figure 19.20b.
(c) More heat is required for the constant pressure process. ΔU is the same in both cases. For constant volume W = 0 and for constant pressure W > 0. The additional heat energy required for constant pressure goes into expansion work.
(d) Q = (0 0100 mol)(12.47 J/mol K)(40.0$^0$C) = 4.99 J
for both processes. ΔU is path independent and for an ideal gas depends only on ΔT.
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