Q#19.63
A monatomic ideal gas expands slowly to twice its original volume, doing 300 J of work in the process. Find the heat added to the gas and the change in internal energy of the gas if the process is
(a) isothermal;
(b) adiabatic;
(c) isobaric.
Answer:
In each case calculate either ΔU or Q for the specific type of process and then apply the first law.
(a) isothermal (ΔT = 0), ΔU = Q − W; W = +300 J.
For any process of an ideal gas, ΔU = n$C_v\Delta T$
Therefore, for an ideal gas, if ΔT = 0 then ΔU = 0 and Q = W= +300 J.
(b) adiabatic (Q = 0). ΔU = Q − W; W = +300 J
Q = 0 says ΔU = −W = −300 J
(c) isobaric Δp = 0. Use W to calculate ΔT and then calculate Q.
W = pΔV = nRΔT
ΔT = W/nR
Q = n$C_p\Delta T$ and for a monatomic ideal gas $C_p=\frac{5}{2}R$.
Thus Q = $n\frac{5}{2}R\Delta T$ = (5Rn/2)(W/nR) = 5W/2 = +750 J
ΔU = n$C_V \Delta T$ for any ideal gas process and $C_V=C_p-R=\frac{3}{2}R$.
Thus ΔU = 3W/2 = +450 J
300 J of energy leaves the gas when it performs expansion work. In the isothermal process this energy is replaced by heat flow into the gas and the internal energy remains the same. In the adiabatic process the energy used in doing the work decreases the internal energy. In the isobaric process 750 J of heat energy enters the gas, 300 J leaves as the work done and 450 J remains in the gas as increased internal energy.
Q#19.64
A cylinder with a piston contains 0.250 mol of oxygen at 2.40 $\times 10^5$ Pa and 355 K. The oxygen may be treated as an ideal gas. The gas first expands isobarically to twice its original volume. It is then compressed isothermally back to its original volume, and finally it is cooled isochorically to its original pressure.
(a) Show the series of processes on a pV-diagram.
(b) Compute the temperature during the isothermal compression.
(c) Compute the maximum pressure.
(d) Compute the total work done by the piston on the gas during the series of processes.
Answer:
pV = nRT. For the isobaric process, W = pΔV = nRΔT. For the isothermal process,
W = nRT ln $\frac{V_f}{V_i}$ and R = 8.315 J/mol.K
(a) The pV diagram for these processes is sketched in Figure 19.64.
(b) Find $T_2$. For process 1 → 2, n, R and p are constant so $\frac{T}{R}=\frac{p}{nR}$ = constant.
$\frac{T_1}{V_1}=\frac{T_2}{V_2}$ and
$T_2=T_1\left(\frac{V_1}{V_1}\right)=(355 \ K)(2)$ = 710 K
(c) The maximum pressure is for state 3. For process 2 → 3, n, R and T are constant. $p_2V_2=p_3V_3$ and
$p_3=p_2\left(\frac{V_2}{V_3}\right)=(2.40 \times 10^5 \ Pa)(2)=4.80 \times 10^5$ Pa
(d) process 1 → 2:
W = $p\Delta V$ = nR$\Delta T$
W = (0.250 mol)(8.315 J/mol.K)(710 K − 355 K) = 738 K
process 2 → 3:
W = nRT ln $\frac{V_3}{V_2}=(0.250 \ mol)(8.215 \ J/mol.K)(710 \ K)ln \left(\frac{1}{2}\right)$
W = −1023 J
process 3 → 1: ΔV = 0 and W = 0.
The total work done is 738 J + (−1023 J) = −285 J. This is the work done by the gas. The work done on the gas is 285 J.
The final pressure and volume are the same as the initial pressure and volume, so the final state is the same as the initial state. For the cycle, ΔU = 0 and Q = W= −285 J. During the cycle, 285 J of heat energy must leave the gas.
Q#19.65
Use the conditions and processes of Problem 19.64 to compute
(a) the work done by the gas, the heat added to it, and its internal energy change during the initial expansion;
(b) the work done, the heat added, and the internal energy change during the final cooling;
(c) the internal energy change during the isothermal compression.
Answer:
Use the ideal gas law, the first law and expressions for Q and W for specific types of processes.
(a) initial expansion (state 1 → state 2)
$p_1=2.40 \times 10^5$ Pa, $T_1$ = 255 K, $p_2=2.40 \times 10^5$ Pa, $ V_2=2V_1$
pV = nRT = ; T/V = pn/R = constant, so
$\frac{T_1}{V_1}=\frac{T_2}{V_2}$ and $T_2=T_2\frac{V_2}{V_1}$
$T_2=(355 \ K)\frac{2V_1}{V_1}$ = 710 K
$\Delta p = 0$ so W = p$\Delta V$ = nR$\Delta T$
W = (0.250 mol)(8.3145 K/mol.K)(710 K $-$ 355 K) = +738 J
Q = $nC_p\Delta T$ = (0.250 mol)(29.17 J/mol.K)(710 K $-$ 355 K) = +2590 J
$\Delta U = Q - W = 2590 \ J - 738 J$ = 1850 J
(b) At the beginning of the final cooling process (cooling at constant volume), T = 710 K. The gas returns to its original volume and pressure, so also to its original temperature of 355 K. ΔV = 0 so W = 0.
Q = $nC_v\Delta T$ = (0.250 mol)(20.85 J/mol.K)(355 K - 710 K) = $-$1850 J
$\Delta U = Q - W = -$1850 J
(c) For any ideal gas process ΔU = $nC_v\Delta T$ For an isothermal process, ΔT = 0 so ΔU = 0.
The three processes return the gas to its initial state, so total Δ = U 0; our results agree with this.
Q#19.66 V
A cylinder with a piston contains 0.150 mol of nitrogen at 1.80 $\times 10^5$ Pa and 300 K. The nitrogen may be treated as an ideal gas. The gas is first compressed isobarically to half its original volume. It then expands adiabatically back to its original volume, and finally it is heated isochorically to its original pressure.
(a) Show the series of processes in a pV-diagram.
(b) Compute the temperatures at the beginning and end of the adiabatic expansion.
(c) Compute the minimum pressure.
Answer:
pV = nRT. For an adiabatic process of an ideal gas,
$T_1V_1^{\gamma - 1}=T_1V_1^{\gamma - 1}$
For $N_1$ = 1.40
(a) The pV-diagram is sketched in Figure 19.66.
(150 K)(1/2)$^{0.4}$ = 114 K.
(c) The minimum pressure occurs at the end of the adiabatic expansion (state 3). During the final heating the volume is held constant, so the minimum pressure is proportional to the Kelvin temperature,
$p_{min}=(1.80 \times 10^5 \ Pa)(114 \ K/300 \ K)=6.82 \times 10^5$ Pa
In the adiabatic expansion the temperature decreases.
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