Q#19.4
Work Done by the Lungs. The graph in Fig. E19.4 shows a pV-diagram of the air in a human lung when a person is inhaling and then exhaling a deep breath. Such graphs, obtained in clinical practice, are normally somewhat curved, but we have modeled one as a set of straight lines of the same general shape. (Important: The pressure shown is the gauge pressure, not the absolute pressure.)
(a) How many joules of net work does this person’s lung do during one complete breath?
(b) The process illustrated here is somewhat different from those we have been studying, because the pressure change is due to changes in the amount of gas in the lung, not to temperature changes. (Think of your own breathing. Your lungs do not expand because they’ve gotten hot.) If the temperature of the air in the lung remains a reasonable 20°C, what is the maximum number of moles in this person’s lung during a breath?
Answer:
The work done in a cycle is the area enclosed by the cycle in a pV diagram.
(a) 1 mm of Hg = 133.3 Pa. $p_{gauge} = p -p_{air}$. In calculating the enclosed area only changes in pressure enter and you can use gauge pressure. 1 L = $10^{-3} \ m^3$.
By counting squares and noting that the area of 1 square is (1 mm of Hg)(0.1 L), we estimate that the area enclosed by the cycle is about 7.5 (mm of Hg). L = 1.00 N.m. The net work done is positive.
(b) Since pV = nRT and T is constant, the maximum number of moles of air in the lungs is when pV is a maximum. In the ideal gas law the absolute pressure p = $p_{gauge}+p_{air}$ must be used.
$p_{air}$ = 760 mm of Hg. 1 mm of Hg = 1 torr.
The maximum pV is when p = 11 torr + 760 torr = 771 torr = 1.028 $\times 10^5$ Pa and
V = 1.4 L = 1.4 $\times 10^{-3} \ m^3$
The maximum pV is
$(pV)_{max}$ = 144 N.m
pV = nRT, so
$n_{max} = \frac{(pV)_{max}}{RT}$
$n_{max} = \frac{144 \ N.m}{(8.3145 \ J/mol.K)(293 \ K)}=0.059$ mol
While inhaling the gas does positive work on the lungs, but while exhaling the lungs do work on the gas, so the net work is positive.
Q#19.5
During the time 0.305 mol of an ideal gas undergoes an isothermal compression at 22$^0$C, 468 J of work is done on it by the surroundings. (a) If the final pressure is 1.76 atm, what was the initial pressure? (b) Sketch a pV-diagram for the process.
Answer:
For an isothermal process W = nRT ln$\left(\frac{p_1}{p_2}\right)$. Solve for $p_1$
For a compression (V decreases) W is negative, so W = -468 J. T = 295.15 K
(a) $\frac{W}{nRT}=ln\left(\frac{p_1}{p_2}\right)$
$\frac{p_1}{p_2}=e^{W/nRT}$
$\frac{W}{nRT}=\frac{-468 \ J}{(0.305 \ mol)(8.3145 \ J/mol.K)(295.15 \ K)}=-0.6253$
So, $p_1=p_2e^{W/nRT}$
$p_1=(1.76 \ atm)e^{-0.6253}=0.942 \ atm$
(b) In the process the pressure increases and the volume decreases. The pV-diagram is sketched in Figure 19.5.
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| Fig.19.5 |
W is the work done by the gas, so when the surroundings do work on the gas, W is negative. The gas was compressed at constant temperature, so its pressure must have increased, which means that $p_1$ < $p_2$, which is what we found.
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