Q#19.67
Use the conditions and processes of Problem 19.66 to compute
(a) the work done by the gas, the heat added to it, and its internal energy change during the initial compression;
(b) the work done by the gas, the heat added to it, and its internal energy change during the adiabatic expansion;
(c) the work done, the heat added, and the internal energy change during the final heating.
Answer:
Use the appropriate expressions for Q, W and ΔU for each type of process. ΔU = Q − W can also be used.
For $N_2$, $C_V$ = 20.76 J/mol.K and $C_p$ = 29.07 J/mol.K
(a) W = pΔV = nRΔT = (0.150 mol)(8.3145 J/mol K)(−150 K) = −187 J,
Q = n$C_p\Delta T$ = (0.150 mol)(20.07 J/mol K)(−150 K) = −654 J,
$\Delta U = Q - W=-467 \ J$
(b) From Eq.
W = $\frac{1}{\gamma - 1}(p_1V_1-p_2V_2)$
using the expression for the temperature found in Problem 19.66,
W = $\frac{1}{\gamma - 1}nR(T_1-T_2)$
because $\frac{T_2}{T_1}=\left(\frac{V_1}{V_2}\right)^{\gamma -1}$, so
W = $\frac{1}{\gamma - 1}nRT_1\left[1-\left(\frac{V_1}{V_2}\right)^{\gamma -1}\right]$
W = $\frac{1}{1.40 - 1}(0.150 \ mol)(8.3145 J/mol.K)(150 K)\left[1-\left(\frac{V_1}{2V_1}\right)^{1.40 -1}\right]$
W = $\frac{1}{0.40}(0.150 \ mol)(8.3145 J/mol.K)(150 K)\left[1-\left(\frac{1}{2}\right)^{0.40}\right]=$ 113 J
Q = 0 for an adiabatic process, and
ΔU = Q − W = −113 J
(c) ΔV = 0, so W = 0. Using the temperature change as found in Problem 19.66 part (b),
Q = $nC_v\Delta T$ = (0.150 mol)(20.76 J/mol.K)(300 K − 113.7 K) = 580 J
ΔU = Q − W = Q = 580 J
For each process we could also use ΔU = $nC_V\Delta T$ to calculate $\Delta U$.
Q#19.68
Comparing Thermodynamic Processes. In a cylinder, 1.20 mol of an ideal monatomic gas, initially at 3.60 $\times 10^5$ Pa and 300 K, expands until its volume triples. Compute the work done by the gas if the expansion is
(a) isothermal;
(b) adiabatic;
(c) isobaric.
(d) Show each process in a pV-diagram. In which case is the absolute value of the work done by the gas greatest? Least?
(e) In which case is the absolute value of the heat transfer greatest? Least?
(f) In which case is the absolute value of the change in internal energy of the gas greatest? Least?
Answer:
Use the appropriate expression for W for each type of process.
For a monatomic ideal gas, γ = 5/3 and $C_V$ = 3R/2.
(a) W = nRT ln ($V_1/V_2$) = nRT ln (3) = $3.29 \times 10^3$ J
(b) Q = 0 so W = $-\Delta U=-nC_V\Delta T$
$\frac{T_2}{T_1}=\left(\frac{V_1}{V_2}\right)^{\gamma -1}$ gives
$T_2=T_1(1/3)^{2/3}$, then
W = $nC_VT_1[1-(1/3)^{2/3}]=2.33 \times 10^3$ J
(c) $V_2=3V_1$, so W = p$\Delta V$ = 2p$V_1$ = 2nR$T_1$ = 6.00 $\times 10^3 \ J$
(d) Each process is shown in Figure 19.68. The most work done is in the isobaric process, as the pressure is maintained at its original value. The least work is done in the adiabatic process.
(e) The isobaric process involves the most work and the largest temperature increase, and so requires the most heat. Adiabatic processes involve no heat transfer, and so the magnitude is zero.
(f) The isobaric process doubles the Kelvin temperature, and so has the largest change in internal energy. The isothermal process necessarily involves no change in internal energy.
The work done is the area under the path for the process in the pV-diagram. Figure 19.68 shows that the work done is greatest in the isobaric process and least in the adiabatic process.
Oscillations of a Piston. A vertical cylinder of radius r contains a quantity of ideal gas and is fitted with a piston with mass m that is free to move (Fig. P19.69). The piston and the walls of the cylinder are frictionless, and the entire cylinder is placed in a constant-temperature bath. The outside air pressure is $p_0$. In equilibrium, the piston sits at a height h above the bottom of the cylinder.
(a) Find the absolute pressure of the gas trapped below the piston when in equilibrium.
(b) The piston is pulled up by a small distance and released. Find the net force acting on the piston when its base is a distance h + y above the bottom of the cylinder, where y is much less than h.
(c) After the piston is displaced from equilibrium and released, it oscillates up and down. Find the frequency of these small oscillations. If the displacement is not small, are the oscillations simple harmonic? How can you tell?
Answer:
At equilibrium the net upward force of the gas on the piston equals the weight of the piston. When the piston moves upward the gas expands, the pressure of the gas drops and there is a net downward force on the piston. For simple harmonic motion the net force has the form , $F_y$ = −ky for a displacement y from equilibrium, and $f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$.
pV = nRT. T is constant.
(a) The difference between the pressure, inside and outside the cylinder, multiplied by the area of the piston, must be the weight of the piston. The pressure in the trapped gas is
$p_0+\frac{mg}{A}=p_0+\frac{mg}{\pi r^2}$
(b) When the piston is a distance h + y above the cylinder, the pressure in the trapped gas is
$\left(p_0+\frac{mg}{\pi r^2}\right)\left(\frac{h}{h+y}\right)$
and for values of y small compared to h,
$\frac{h}{h+y}=\left(1+\frac{y}{h}\right)^{-1} \approx 1-\frac{y}{h}$
The net force, taking the positive direction to be upward, is then
$F_y=\left[\left(p_0+\frac{mg}{\pi r^2}\right)\left(1-\frac{y}{h}\right)-p_0\right](\pi r^2)-mg=-\left(\frac{y}{h}\right)(p_0\pi r^2+mg)$
This form shows that for positive h, the net force is down; the trapped gas is at a lower pressure than the equilibrium pressure, and so the net force tends to restore the piston to equilibrium.
(c) The angular frequency of small oscillations would be given by
$\omega^2=\frac{(p_0\pi r^2+mg)/h}{m}$
$\omega^2=\frac{g}{h}\left(1+\frac{p_0\pi r^2}{mg}\right)$
$\omega=\sqrt{\frac{g}{h}\left(1+\frac{p_0\pi r^2}{mg}\right)}$
because $\omega = 2\pi f$, → $f = \frac{\omega}{2\pi}$, so
$f=\frac{1}{2\pi}\sqrt{\frac{g}{h}\left(1+\frac{p_0\pi r^2}{mg}\right)}$
If the displacements are not small, the motion is not simple harmonic. This can be seen be considering what happens if y ~ −h; the gas is compressed to a very small volume, and the force due to the pressure of the gas would become unboundedly large for a finite displacement, which is not characteristic of simple harmonic motion. If y >> h (but not so large that the piston leaves the cylinder), the force due to the pressure of the gas becomes small, and the restoring force due to the atmosphere and the weight would tend toward a constant, and this is not characteristic of simple harmonic motion.
The assumption of small oscillations was made when $\frac{h}{h+y}$ was replaced by 1 − y/h this is accurate only when y/h is small.
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