Q#19.32
The engine of a Ferrari F355 F1 sports car takes in air at 20.0$^0C$ and 1.00 atm and compresses it adiabatically to 0.0900 times the original volume. The air may be treated as an ideal gas with $\gamma$ = 1.4.
(a) Draw a pV-diagram for this process.
(b) Find the final temperature and pressure.
Answer:
(a) In the process the pressure increases and the volume decreases. The pV-diagram is sketched in Figure 19.32.
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| Fig.19.32 |
For an adiabatic process of an ideal gas,
$p_1V_1^{\gamma}$ = $p_2V_2^{\gamma}$ (*)
$T_1V_1^{\gamma-1}$ = $T_2V_2^{\gamma-1}$ (**) and
pV = nRT
From the first equation, $T_2=T_1(V_1/V_2)^{\gamma-1}$
$T_2=(293 \ K)(V_1/0.0900V_1)^{1.4-1}$
$T_2=(293 \ K)(11.11)^{0.4}$
$T_2= 768 \ K = 495 \ ^0C$
(Note: In the equation $T_1V_1^{\gamma-1}$ = $T_2V_2^{\gamma-1}$ the temperature must be in kelvins.)
$p_1V_1^{\gamma}$ = $p_2V_2^{\gamma}$ implies
$p_2= p_1(V_1/V_2)^{\gamma}$
$p_2= (1.00 \ atm)(V_1/0.0900 V_1)^{1.4}$
$p_2=(1.00 \ atm)(11.11)^{1.4}=29.1 \ atm$
Q#19.33
During an adiabatic expansion the temperature of 0.450 mol of argon (Ar) drops from 50.0$^0C$ to 10.0$^0C$. The argon may be treated as an ideal gas.
(a) Draw a pV-diagram for this process.
(b) How much work does the gas do?
(c) What is the change in internal energy of the gas?
Answer:
(a) In the expansion the pressure decreases and the volume increases. The pV-diagram is sketched in Figure 19.33.
(b) Adiabatic means Q = 0. Then ΔU = Q − W gives
W = $-\Delta U=-nC_v\Delta T = -nC_v(T_2-T_1)$
For Argon $C_v$ = 12.47 J/mol.K, so
W = $-(0.450 \ mol)(12.47 \ J/mol.K)(50.0^0C-10.0^0C)$ = +224 J
W positive for ΔV > 0 (expansion)
(c) $\Delta U = -W=-224 \ J$
There is no heat energy input. The energy for doing the expansion work comes from the internal energy of the gas, which therefore decreases. For an ideal gas, when T decreases, U decreases.
Q#19.34
A player bounces a basketball on the floor, compressing it to 80.0% of its original volume. The air (assume it is essentially gas) inside the ball is originally at a temperature of 20.0$^0C$ and a pressure of 2.00 atm. The ball’s inside diameter is 23.9 cm.
(a) What temperature does the air in the ball reach at its maximum compression? Assume the compression is adiabatic and treat the gas as ideal.
(b) By how much does the internal energy of the air change between the ball’s original state and its maximum compression?
Answer:
Assume the expansion is adiabatic, $T_1V_1^{\gamma-1}$ = $T_2V_2^{\gamma-1}$ relates V and T. Assume the air behaves as an ideal gas, so
$\Delta U = nC_v\Delta T$. Use pV = nRT to calculate n.
Given: For air, $C_v=29.76 \ J/mol.K$, $\gamma = 1.4$, $V_2=0.800V_1$, $T_1=293.15 \ K$, $p_1=2.026 \times 10^5 \ Pa$. For a sphere, $V= \frac{4}{3}\pi r^3$
(a) $T_2=T_1(V_1/V_2)^{\gamma-1}$
$T_2=(293.15 \ K)(V_1/0.800V_1)^{1.4-1}$
$T_2=(293.15 \ K)(1.25)^{0.4}$
$T_2= 320.5 \ K = 47.4 \ ^0C$
(b) $V= \frac{4}{3}\pi r^3=\frac{4\pi}{2}(0.1195 \ m^3)=7.15 \times 10^{-3} \ m^3$
$n = \frac{pV}{RT}=\frac{(2.026 \times 10^5 \ Pa)(7.15 \times 10^{-3} \ m^3)}{(8.314 \ J/mol.K)(293.15 \ K)}$
n = 0.594 mol
$\Delta U = nC_v\Delta T = (0.594 \ mol)(20.76 \ J/mol.K)(321 \ K - 293 \ K)$
$\Delta U = 345 \ J$
Q#19.35
On a warm summer day, a large mass of air (atmospheric pressure 1.01 $\times 10^5$ Pa) is heated by the ground to a temperature of 26$^0C$ and then begins to rise through the cooler surrounding air. (This can be treated approximately as an adiabatic process; why?) Calculate the temperature of the air mass when it has risen to a level at which atmospheric pressure is only 0.850 $\times 10^5$ Pa. Assume that air is an ideal gas, with $\gamma$ = 1.4. (This rate of cooling for dry, rising air, corresponding to roughly $1^0C$ per 100 m of altitude, is called the dry adiabatic lapse rate.)
Answer:
Combine $T_1V_1^{\gamma-1}$ = $T_2V_2^{\gamma-1}$ with PV = nRT to obtain an expression relating T and p for an adiabatic process of an ideal gas.
Given: $T_1=(26+273.15) \ K = 299.15 \ K$, $p_1$ = 1.01 $\times 10^5$ Pa and $p_2$ = 0.850 $\times 10^5$ Pa
$V = \frac{nRT}{p}$ so,
$T_1\left(\frac{nRT_1}{p_1}\right)^{\gamma-1}$ = $T_2\left(\frac{nRT_2}{p_2}\right)^{\gamma-1}$
$T_1\left(\frac{T_1}{p_1}\right)^{\gamma-1}$ = $T_2\left(\frac{T_2}{p_2}\right)^{\gamma-1}$
$T_2=T_1\left(\frac{p_2}{p_1}\right)^{\frac{\gamma-1}{\gamma}}$
$T_2=(299.15 \ K)\left(\frac{0.850 \times 10^5 \ Pa}{1.01 \times 10^5 \ Pa}\right)^{\frac{1.4-1}{1.4}}$
$T_2=284.8 \ K = 11.6^0C$
For an adiabatic process of an ideal gas, when the pressure decreases the temperature decreases.
Q#19.36
A cylinder contains 0.100 mol of an ideal monatomic gas. Initially the gas is at a pressure of 1.00 $\times 10^5$ Pa and occupies a volume of 2.50 $\times 10^{-3} \ m^3$.
(a) Find the initial temperature of the gas in kelvins.
(b) If the gas is allowed to expand to twice the initial volume, find the final temperature (in kelvins) and pressure of the gas if the expansion is (i) isothermal; (ii) isobaric; (iii) adiabatic.
Answer:
pV = nRT. For an adiabatic process, $T_1V_1^{\gamma-1}$ = $T_2V_2^{\gamma-1}$ (*)
For an ideal monatomic gas, $\gamma = \frac{5}{3}$
(a) $T = \frac{PV}{nR}=\frac{(1.00 \times 10^5 \ Pa)(2.50 \times 10^{-3} \ m^3)}{(0.1 \ mol)(8.3145 \ J/mol.K)}=301 \ K$
(b) (i) Isothermal: If the expansion is isothermal, the process occurs at constant temperature and the final temperature is the same as the initial temperature, namely 301 K.
$p_2=p_1(V_1/V_2)=\frac{1}{2}V_1=5.00 \times 10^4 \ Pa$
(ii) Isobaric: $\Delta p = 0$ so $p_2=1.00 \times 10^5 \ Pa$
$T_2=T_1(V_2/V_1)=2T_1=2(301 \ K)=602 \ K$
(iii) Adiabatic: Using Eq.(*)
$T_2=T_1(V_1/V_2)^{\gamma-1}$
$T_2=(301 \ K)(V_1/2V_1)^{1.67-1}$
$T_2=(301 \ K)(1/2)^{0.67} = 189 \ K$
Then pV = nRT gives $p_2 = 3.14 \times 10^4$ Pa.
In an isobaric expansion, T increases. In an adiabatic expansion, T decreases.
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