Q#19.53
In a certain process, of heat is liberated by a system, and at the same time the system contracts under a constant external pressure of The internal energy of the system is the same at the beginning and end of the process. Find the change in volume of the system. (The system is not an ideal gas.)
Answer:
Use the first law to calculate W and then use W = pΔV for the constant pressure process to calculate ΔV.
Known: Q = −2.15 × 10$^5$ J (negative since heat energy goes out of the system)
ΔU = 0 so W = Q = −2.15 × 10$^5$ J
Constant pressure, so
W = pΔV
Then $\Delta V = \frac{-2.15 \times 10^5 \ J}{9.50 \times 10^5 \ Pa}=-0.226 \ m^3$
Positive work is done on the system by its surroundings; this inputs to the system the energy that then leaves the system as heat.
Q#19.54
A cylinder with a frictionless, movable piston like that shown in Fig. 19.5 contains a quantity of helium gas. Initially the gas is at a pressure of has a temperature of 300 K, and occupies a volume of 1.50 L. The gas then undergoes two processes. In the first, the gas is heated and the piston is allowed to move to keep the temperature equal to 300 K. This continues until the pressure reaches In the second process, the gas is compressed at constant pressure until it returns to its original volume of 1.50 L. Assume that the gas may be treated as ideal.
(a) In a pV-diagram, show both processes.
(b) Find the volume of the gas at the end of the first process, and find the pressure and temperature at the end of the second process.
(c) Find the total work done by the gas during both processes.
(d) What would you have to do to the gas to return it to its original pressure and temperature?
Answer:
pV = nRT. For an isothermal process W = nRT ln $\left(\frac{V_2}{V_1}\right)$. For a constant pressure process, W = pΔV.
(a) The pV-diagram is sketched in Figure 19.54.
$V_2=V_1(p_1/p_2)$
$V_2=(1.5 \ L)[(1.00 \times 10^5 \ Pa)/(2.50 \times 10^5 \ Pa)]=6.00 \ L$.
The final pressure is given as being the same as
$p_3=p_2=2.5 \times 10^5$ Pa
The final volume is the same as the initial volume, so
$T_3=T_1(p_3/p_1)=75.0 \ K$
(c) Treating the gas as ideal, the work done in the first process is
W = nRT ln $\left(\frac{V_2}{V_1}\right)=p_1V_1 ln \left(\frac{p_1}{p_2}\right)$
W = $(1.00 \times 10^5 \ Pa)(1.5 \times 10^{-3} \ m^3)ln \left(\frac{1.00 \times 10^5 \ Pa}{2.50 \times 10^5 \ Pa}\right)=208 \ J$
For the second process,
W = $p_2(V_3-V_2)=p_2(V_1-V_2)=p_2V_1\left(1-\frac{p_1}{p_2}\right)$
$W = (2.50 \times 10^4 \ Pa)(1.5 \times 10^{-3} \ m^3)\left(1-\frac{1.00 \times 10^5 \ Pa}{2.50 \times 10^5 \ Pa}\right)=-113 \ J$
The total work done is 208 J − 113 J = 95 J
(d) Heat at constant volume. No work would be done by the gas or on the gas during this process.
When the volume increases, W > 0. When the volume decreases, W < 0.
Q#19.55
A Thermodynamic Process in a Liquid. A chemical engineer is studying the properties of liquid methanol She uses a steel cylinder with a cross-sectional area of and containing of methanol. The cylinder is equipped with a tightly fitting piston that supports a load of The temperature of the system is increased from to For methanol, the coefficient of volume expansion is the density is and the specific heat at constant pressure is You can ignore the expansion of the steel cylinder. Find
(a) the increase in volume of the methanol;
(b) the mechanical work done by the methanol against the force;
(c) the amount of heat added to the methanol;
(d) the change in internal energy of the methanol.
(e) Based on your results, explain whether there is any substantial difference between the specific heats (at constant pressure) and (at constant volume) for methanol under these conditions.
Answer:
ΔV = $V_0 \beta \Delta T$, W = pΔV since the force applied to the piston is constant. Q = m$c_p$ΔT and ΔU = Q − W.
(a) The change in volume is
ΔV = $V_0 \beta \Delta T=(1.2 \times 10^{-2} \ m^3)(1.20 \times 10^{-3} \ K^{-1})(30.0 \ K)=4.32 \times 10^{-4} \ m^3$
(b) W = pΔV = $(F/A)\Delta V = \left(\frac{3.00 \times 10^4 \ N}{0.0200 \ m^2}\right)(4.32 \times 10^{-4} \ m^3)=648 \ J$
(c) Q = m$c_p$ΔT = $V_0 \rho c_p \Delta T=(1.20 \times 10^{-2} \ m^3)(791 \ kg/m^3)(2.51 \times 10^3 \ J/kg.K)(30.0 \ K)$
Q = $7.15 \times 10^5$ J
(d) ΔU = Q − W = $7.15 \times 10^5$ J to three figures.
(e) Under these conditions W is much less than Q and there is no substantial difference between $c_v$ and $c_p$.
Q#19.56
A Thermodynamic Process in a Solid. A cube of copper 2.00 cm on a side is suspended by a string. (The physicalproperties of copper are given in Tables 14.1, 17.2, and 17.3.) The cube is heated with a burner from 20.0$^0C$ to 90.0$^0C$. The air surrounding the cube is at atmospheric pressure (1.00 $\times 10^5$ Pa). Find
(a) the increase in volume of the cube;
(b) the mechanical work done by the cube to expand against the pressure of the surrounding air;
(c) the amount of heat added to the cube;
(d) the change in internal energy of the cube.
(e) Based on your results, explain whether there is any substantial difference between the specific heats $c_p$ (at constant pressure) and $c_v$ (at constant volume) for copper under these conditions.
Answer:
ΔV = $V_0 \beta \Delta T$, W = pΔV since the force applied to the piston is constant. Q = m$c_p$ΔT and ΔU = Q − W.
For copper, $\beta = 5.1 \times 10^{-5}/^0C$, $c_p$ = 390 J/kg.K and $\rho$ = 8.90 $\times 10^3 \ kg/m^3$
(a) ΔV = $V_0 \beta \Delta T=(2.0 \times 10^{-2} \ m)^3(5.1 \times 10^{-5} \ ^0C^{-1})(70.0 \ ^0C)=4.32 \times 10^{-4} \ m^3$
ΔV = $2.86 \times 10^{-8} \ m^3$
(b) W = pΔV = 2.88 $\times 10^{-3}$ J
(c) Q = m$c_p$ΔT = $V_0 \rho c_p \Delta T=(8.00 \times 10^{-6} \ m^3)(8.9 \times 10^3 \ kg/m^3)(390 \ J/kg.K)(70.0 \ ^0C)$
Q = 1944 J
(d) To three figures, ΔU = Q = 1944 J.
(e) Under these conditions, the difference is not substantial, since W is much less than Q.
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