Q#19.49
(a) One-third of a mole of He gas is taken along the path abc shown as the solid line in Fig. P19.49. Assume that the gas may be treated as ideal. How much heat is transferred into or out of the gas?
(b) If the gas instead went from state a to state c along the horizontal dashed line in Fig. P19.49, how much heat would be transferred into or out of the gas?
(c) How does Q in part (b) compare with Q in part (a)? Explain
Answer:
The segments ab and bc are not any of the familiar ones, such as isothermal, isobaric or isochoric, but ac is isobaric.
For helium, $C_v$ = 12.47 J/mol.K and $C_p$ = 20.78 J/mol.K. ΔU = Q − W. W is the area under the p versus V curve. ΔU = n$C_v$ΔT for any process of an ideal gas.
(a) W = $\frac{1}{2}(1.0 \times 10^5 \ Pa + 3.5 \times 10^5 \ Pa)(0.0060 \ m^3-0.0020 \ m^3)$
$+\frac{1}{2}(1.0 \times 10^5 \ Pa + 3.5 \times 10^5 \ Pa)(0.0100 \ m^3-0.0060 \ m^3)=1800 \ J$
Find $\Delta T = T_c-T_a$. p is constant so
$\Delta T=\frac{p\Delta V}{nR}$
$\Delta T=\frac{(1.0 \times 10^5 \ Pa)(0.0100 \ m^3-0.0020 \ m^3)}{(1/3 \ mol)(8.3145 \ J/mol.K)}$
$\Delta T =$ 289 K
ΔU = n$C_v$ΔT = (1/3 mol)(12.47 J/mol.K)(289 K)
ΔU = 1.20 $\times 10^3$ J
Q = ΔU + W = 1.20 $\times 10^3$ J + 1800 J = 3.00 $\times 10^3$ J
Q > 0, so this heat is transferred into the gas.
(b) This process is isobaric, so
Q = n$C_p$ΔT = (1/3 mol)(20.78 J/mol.K)(289 K) = 2.00 $\times 10^3$ J
Q > 0, so this heat is transferred into the gas.
(c) Q is larger in part (a).
ΔU is the same in parts (a) and (b) because the initial and final states are the same, but in (a) more work is done.
Q#19.50
Two moles of helium are initially at a temperature of 27.0°C and occupy a volume of 0.0300 $m^3$. The helium first expands at constant pressure until its volume has doubled. Then it expands adiabatically until the temperature returns to its initial value. Assume that the helium can be treated as an ideal gas.
(a) Draw a diagram of the process in the pV-plane.
(b) What is the total heat supplied to the helium in the process?
(c) What is the total change in internal energy of the helium?
(d) What is the total
work done by the helium?
(e) What is the final volume of the
helium?
Answer:
We have an isobaric expansion followed by an adiabatic expansion.
$T_1$ = 300 K. When the volume doubles at constant pressure the temperature doubles, so $T_2$ = 600 K. For helium, $C_p$ = 20.78 J/mol.K and γ = 1.67.
ΔU = n $C_v$ΔT for any process of an ideal
gas. ΔU = Q − W.
(a) The process is sketched in Figure 19.50.
(b) For the isobaric step,
Q = n$C_p$ΔT = (2.00 mol)(20.78 J/mol K)(300 K) = 1.25 $\times 10^4$J.
For the adiabatic process, Q = 0.
The total heat is Q is 1.25 $\times 10^4$ J.
(c) ΔU = 0 since ΔT = 0
(d) Since ΔU = 0, W = Q = 1.25 × 10$^4$ J.
(e) $T_1$ = 300 K, $T_2$ = 600 K and $V_2$ = 0.0600 m$^3$ .
$T_2V_2^{\gamma - 1} = T_3V_3^{\gamma - 1}$
$V_3=V_2\left(\frac{T_2}{T_3}\right)^{1/(\gamma - 1)}$
$V_3=(0.00600 \ m^3)\left(\frac{600 \ K}{300 \ K}\right)^{1/(0.67)}=0.169 \ m^3$
In both processes the internal energy changes. In the isobaric expansion the temperature
increases and the internal energy increases. In the adiabatic expansion the temperature decreases and ΔU < 0.
The magnitudes of the two temperature changes are equal and the net change in internal energy is zero.
Q#19.51
Starting with 2.50 mol of gas (assumed to be ideal) in a cylinder at 1.00 atm and 20.0°C a chemist first heats the gas at constant volume, adding 1.52 $\times 10^4$ J of heat, then continues heating and allows the gas to expand at constant pressure to twice its original volume.
(a) Calculate the final temperature of the gas.
(b) Calculate the amount of work done by the gas.
(c) Calculate the amount of heat added to the gas while it was expanding.
(d) Calculate the change in internal energy of the gas for the whole process.
Answer:
Use Q = n$C_v$ΔT to calculate the temperature change in the constant volume process and use
pV nRT = to calculate the temperature change in the constant pressure process.
The work done in the constant volume process is zero and the work done in the constant pressure process is W = pΔV.
Use Q = n$C_p$ΔT to calculate the heat flow in the constant pressure process. ΔU = n$C_v$ΔT or ΔU = Q − W.
For $N_2$ $C_v$ = 20.76 J/mol.K and $C_p$ = 29.07 J/mol.K
(a) For process ab,
$\Delta T = \frac{Q}{nC_v}=\frac{1.52 \times 10^4 \ J}{(2.50 \ mol)(20.76 \ J/mol.K)}=293 \ K$
$T_a=293 \ K$, so $T_b=586 \ K$
pV nRT = says T doubles when V doubles and p is constant, so
$T_c=2(586 \ K)=1172 \ K = 899^0C$
(b) For process ab, $W_{ab}=0$. For process bc,
$W_{bc}=p \Delta V= nR \Delta T=(2.50 \ mol)$(8.314 J/mol.K)(1172 K - 586 K) = 1.22 $\times 10^4$ J.
$W=W_{ab}+W_{bc}=1.22 \times 10^4$ J
(c) For process bc,
Q = n$C_p\Delta T$ = (2.50 mol)(29.07 J/mol.K)(1172 K - 586 K) = 4.26 $\times 10^4$ J.
(d) $\Delta U = nC_v\Delta T$ = (2.50 mol)(20.76 J/mol.K)(1172 K - 586 K) = 4.56 $\times 10^4$ J.
Q#19.52
Nitrogen gas in an expandable container is cooled from 50.0°C to 10.0°C with the pressure held constant at 3.00 $\times 10^5$ Pa. The total heat liberated by the gas is 2.50 $\times 10^4$ J. Assume that the gas may be treated as ideal.
(a) Find the number of moles of gas.
(b) Find the change in internal energy of the gas.
(c) Find the work done by the gas.
(d) How much heat would be liberated by the gas for the same temperature change if the volume were constant?
Answer:
For a constant pressure process, Q = n$C_p$ΔT, ΔU = Q − W and ΔU = n$C_v$ΔT for any ideal gas process.
For $N_2$ $C_v$ = 20.76 J/mol.K and $C_p$ = 29.07 J/mol.K
(a) $n = \frac{Q}{C_p\Delta T}=\frac{-2.5 \times 10^4 \ J}{(29.07 \ J/mol.K)(-40.0 \ K)}=21.5 \ mol$
(b) ΔU = n$C_v$ΔT = $Q(C_v/C_p) = (-2.5 \times 10^4 \ J)(20.76/29.07)=-1.79 \times 10^4 \ J$
(c) W = Q $-\Delta U=-7.15 \times 10^3$ J
(d) ΔU is the same for both processes, and if 0, ΔV = 0, W = 0 and $Q = \Delta U=-1.79 \times 10^4$ J
For a given ΔT, Q is larger in magnitude when the pressure is constant than when the volume is constant.
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