Q#19.60
A certain ideal gas has molar heat capacity at constant volume $C_V$. A sample of this gas initially occupies a volume $V_0$ at pressure $p_0$ and absolute temperature $T_0$. The gas expands isobarically to a volume 2$V_0$ and then expands further adiabatically to a final volume 4$V_0$
(a) Draw a pV-diagram for this sequence of processes.
(b) Compute the total work done by the gas for this sequence of processes.
(c) Find the final temperature of the gas.
(d) Find the absolute value |Q| of the total heat flow into or out of the gas for this sequence of processes, and state the direction of heat flow.
Answer:
For constant pressure, W = pΔV. For an adiabatic process of an ideal gas,
W = $\frac{C_V}{R}(p_1V_1-p_2V_2)$ and $p_1V_1^{\gamma}=p_2V_2^{\gamma}$
$\gamma = \frac{C_p}{C_V}=\frac{C_V+R}{C_V}=1+\frac{R}{C_V}$
(a) The pV-diagram is sketched in Figure 19.60.
(b) The work done is
W = $p_0(2V_0-V_0)+\frac{C_V}{R}[p_0(2V_0)-p_3(4V_0)]$
because $p_3=p_0(2V_0/4V_0)^{\gamma}=p_0(2^{-\gamma})$ and so
W = $p_0V_0\left[1+\frac{C_V}{R}(2-2^{2-\gamma})\right]$
Note that $p_0$ is the absolute pressure.
(c) The most direct way to find the temperature is to find the ratio of the final pressure and volume to the original and treat the air as an ideal gas.
$p_3=p_2\left(\frac{V_2}{V_3}\right)^{\gamma}=p_1\left(\frac{V_2}{V_3}\right)^{\gamma}$
since $p_1=p_2$. Then
$T_3=T_0\frac{p_3V_3}{p_1V_1}=T_0\left(\frac{V_2}{V_3}\right)^{\gamma}\left(\frac{V_3}{V_1}\right)$
$T_3=T_0\left(\frac{1}{2}\right)^{\gamma} .4=T_02^{2-\gamma}$
(d) Since $n=\frac{p_0V_0}{RT_0}$
Q = $\frac{p_0V_0}{RT_0}(C_V+R)(2T_0-T_0)=p_0V_0\left(\frac{C_V}{R}+1\right)$
This amount of heat flows into the gas, since Q > 0.
Q#19.61
An air pump has a cylinder 0.250 m long with a movable piston. The pump is used to compress air from the atmosphere (at absolute pressure 1.01 $\times 10^5$ Pa) into a very large tank at gauge pressure 4.30 $\times 10^5$ Pa. (For air, $C_V$ = 20.8 J/mol.K)
(a) The piston begins the compression stroke at the open end of the cylinder. How far down the length of the cylinder has the piston moved when air first begins to flow from the cylinder into the tank? Assume that the compression is adiabatic.
(b) If the air is taken into the pump at 27.0 $^0C$ what is the temperature of the compressed air?
(c) How much work does the pump do in putting 20.0 mol of air into the tank?
Answer:
(a) $\gamma = \frac{C_p}{C_V}=\frac{C_V+R}{C_V}=1+\frac{R}{C_V}=1.40$
The two positions of the piston are shown in Figure 19.61.
$p_1=1.01 \times 10^5$ Pa
$p_2=4.20 \times 10^5 \ Pa+p_{air}=5.21 \times 10^5$ Pa
$V_1=h_1A$ and $V_2=h_2A$
adiabatic process: $p_1V_1^{\gamma}=p_2V_2^{\gamma}$
$p_1h_1^{\gamma}A^{\gamma}=p_2h_2^{\gamma}A^{\gamma}$
$h_2=h_1\left(\frac{p_1}{p_2}\right)^{1/\gamma}$
$h_2=(0.250 \ m)\left(\frac{1.01 \times 10^5 \ Pa}{5.21 \times 10^5 \ Pa}\right)^{1/1.40}=0.0774 \ m$
The piston has moved a distance
$h_1-h_2=0.250 \ m - 0.0774 \ m=0.173 \ m$
(b) $T_1V_1^{\gamma-1}=T_2V_2^{\gamma-1}$
$T_1h_1^{\gamma-1}A^{\gamma-1}=T_2h_2^{\gamma-1}A^{\gamma-1}$
$T_2=T_1\left(\frac{h_1}{h_2}\right)^{\gamma-1}$
$T_2=(300.1 \ K)\left(\frac{0.250 \ m}{0.0774 \ m}\right)^{1.40-1}=479.7 \ K=207^0C$
(c) W = $nC_V(T_1-T_2)$
W = $(20.0 \ mol)(20.8 \ J/mol.K)(300.1 \ K - 479.7 \ K)=-7.47 \times 10^4$ J
In an adiabatic compression of an ideal gas the temperature increases. In any compression the work W is negative.
Q#19.62
Engine Turbochargers and Intercoolers. The power output of an automobile engine is directly proportional to the mass of air that can be forced into the volume of the engine’s cylinders to react chemically with gasoline. Many cars have a turbocharger, which compresses the air before it enters the engine, giving a greater mass of air per volume. This rapid, essentially adiabatic compression also heats the air. To compress it further, the air then passes through an intercooler in which the air exchanges heat with its surroundings at essentially constant pressure. The air is then drawn into the cylinders. In a typical installation, air is taken into the turbocharger at atmospheric pressure (1.01 $\times 10^5$ Pa) density $\rho$ = 1.23 kg/$m^3$ and temperature 15.0$^0C$. It is compressed adiabatically to 1.45 $\times 10^5$ Pa. In the intercooler, the air is cooled to the original temperature of 15.0$^0C$at a constant pressure of 1.45 $\times 10^5$ Pa.
(a) Draw a pV-diagram for this sequence of processes.
(b) If the volume of one of the engine’s cylinders is 575 $cm^3$ what mass of air exiting from the intercooler will fill the cylinder at 1.45 $\times 10^5$ Pa? Compared to the power output of an engine that takes in air at 1.01 $\times 10^5$ Pa at 15.0$^0C$ what percentage increase in power is obtained by using the turbocharger and intercooler?
(c) If the intercooler is not used, what mass of air exiting from the turbocharger will fill the cylinder at 1.45 $\times 10^5$ Pa? Compared to the power output of an engine that takes in air at 1.01 $\times 10^5$ Pa at 15.0$^0C$ what percentage increase in power is obtained by using the turbocharger alone?
Answer:
m = ρV. The density of air is given by $\rho=\frac{pM}{RT}$
For an adiabatic process, $T_1V_1^{\gamma-1}=T_2V_2^{\gamma-1}$ gives
$T_1p_1^{1-\gamma}=T_2p_2^{1-\gamma}$
(a) The pV-diagram is sketched in Figure 19.62.
$m=\rho V \frac{p}{p_{air}}=(1.23 \ kg/m^3)(575 \times 10^{-6} \ m^3)\frac{1.45 \times 10^5 \ Pa}{1.01 \times 10^5 \ Pa}=1.02 \times 10^{-3} \ kg$
Without the turbocharger or intercooler the mass of air at T = 15.0$^0C$ and $p=1.01 \times 10^5$ Pa in a cylinder is m = $\rho_0$V = 7.07 $\times 10^{-4}$ kg. The increase in power is proportional to the increase in mass of air in the cylinder; the percentage increase is
$\frac{1.02 \times 10^{-3} \ kg}{7.07 \times 10^{-4} \ kg}-1=0.44 = 44$%
(c) The temperature after the adiabatic process is
$T_2=T_1\left(\frac{p_2}{p_1}\right)^{(\gamma-1)/\gamma}$.
The density becomes
$\rho = \rho_0\left(\frac{T_1}{T_2}\right)\left(\frac{p_2}{p_1}\right)$
$\rho = \rho_0\left(\frac{p_2}{p_1}\right)^{(1-\gamma)/\gamma}\left(\frac{p_1}{p_1}\right)$
$\rho = \rho_0\left(\frac{p_2}{p_1}\right)^{1/\gamma}$
The mass of air in the cylinder is
$m=(1.23 \ kg/m^3)(575 \times 10^{-6} \ m^3)\left(\frac{1.45 \times 10^5 \ Pa}{1.01 \times 10^5 \ Pa}\right)^{1/1.40}=9.16 \times 10^{-4} \ kg$
The percentage increase in power is
$\frac{9.16 \times 10^{-4} \ kg}{7.07 \times 10^{-4} \ kg}-1=0.3 = 30$%
The turbocharger and intercooler each have an appreciable effect on the engine power.
Post a Comment for "The First Law of Thermodynamics Problems and Solutions 8"