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Q 5.41

Expert-verified
Found in: Page 176

### An Introduction to Thermal Physics

Book edition 1st
Author(s) Daniel V. Schroeder
Pages 356 pages
ISBN 9780201380279

# Suppose you have a liquid (say, water) in equilibrium with its gas phase, inside some closed container. You then pump in an inert gas (say, air), thus raising the pressure exerted on the liquid. What happens? (a) For the liquid to remain in diffusive equilibrium with its gas phase, the chemical potentials of each must change by the same amount: $d{\mu }_{l}=d{\mu }_{g}$ Use this fact and equation 5.40 to derive a differential equation for the equilibrium vapour pressure, Pv as a function of the total pressure P. (Treat the gases as ideal, and assume that none of the inert gas dissolves in the liquid.) (b) Solve the differential equation to obtain ${P}_{v}\left(P\right)-{P}_{v}\left({P}_{v}\right)={e}^{\left(P-{P}_{v}\right)V/NkT}$where the ratio V/N in the exponent is that of the liquid. (The term Pv(Pv) is just the vapour pressure in the absence of the inert gas.) Thus, the presence of the inert gas leads to a slight increase in the vapour pressure: It causes more of the liquid to evaporate. (c) Calculate the percent increase in vapour pressure when air at atmospheric pressure is added to a system of water and water vapour in equilibrium at 25°C. Argue more generally that the increase in vapour pressure due to the presence of an inert gas will be negligible except under extreme conditions.

The vapour pressure is 0.07 percent higher than it would be in the absence of air.

See the step by step solution

## Part (a) Step 1: Explanation

Using the equation 5.40, the chemical potential for the gas is as follows.

${\mu }_{g}=\mu °+kT\mathrm{ln}\left({P}_{v}\right)$

Where

$\mu °$ is constant

k is Boltzmann's constant

T is temperature

P is the partial pressure

Differentiating the equation with respect to P

$\frac{d}{dP}\left({\mu }_{g}\right)=\frac{d}{dP}\left(\mu °+kT\mathrm{ln}\left({P}_{v}\right)\right)\phantom{\rule{0ex}{0ex}}=0+kT\left(\frac{1}{{P}_{v}}\right)\frac{d{P}_{v}}{dP}\phantom{\rule{0ex}{0ex}}\frac{d{\mu }_{g}}{dP}=\left(\frac{kT}{{P}_{v}}\right)\frac{d{P}_{v}}{dP}\phantom{\rule{0ex}{0ex}}$

The chemical potential for liquid is

${\mu }_{1}=\frac{G}{N}$

Where G is Gibbs free energy

Differentiating the equation with respect to pressure P

$\frac{d}{dP}\left({\mu }_{l}\right)=\frac{d}{dP}\left(\frac{G}{N}\right)\phantom{\rule{0ex}{0ex}}\frac{d{\mu }_{l}}{dP}=\frac{1}{N}\frac{dG}{dP}$

The volume V is equal to the differentiation of the Gibbs free energy with respect to pressure.

$\frac{dG}{dP}=V\phantom{\rule{0ex}{0ex}}$

Substitute V for $\frac{dG}{dP}$

$\frac{d{\mu }_{l}}{dP}=\frac{1}{N}\left(V\right)=\frac{V}{N}\phantom{\rule{0ex}{0ex}}$

## Step 2: Calculations

The change in chemical potentials for the two phases, liquid and gas, must be equal in equilibrium.

$d{\mu }_{l}=d{\mu }_{g}\phantom{\rule{0ex}{0ex}}\frac{d{\mu }_{l}}{dP}=\frac{d{\mu }_{\mathrm{g}}}{dP}\phantom{\rule{0ex}{0ex}}\text{Substitute values of}\frac{d{\mu }_{1}}{dP}\text{and}\frac{d{\mu }_{g}}{dP}\text{.}\phantom{\rule{0ex}{0ex}}\frac{V}{N}=\left(\frac{kT}{{P}_{v}}\right)\frac{d{P}_{v}}{dP}\phantom{\rule{0ex}{0ex}}\frac{d{P}_{v}}{dP}=\left(\frac{V}{NkT}\right){P}_{v}\phantom{\rule{0ex}{0ex}}$

Hence, the differential equation is $\left(\frac{V}{NkT}\right){P}_{v}$

## Part (b) Step 3: Explanation

$\text{The solution for the differential equation}\frac{d{P}_{v}}{dP}=\left(\frac{V}{NkT}\right){P}_{v}\text{is in exponential form.}\phantom{\rule{0ex}{0ex}}{P}_{v}\left(P\right)=\left(\text{constant}\right){e}^{\frac{PV}{NRT}}\dots \dots .\left(1\right)\phantom{\rule{0ex}{0ex}}\text{Substitute}{P}_{v}\text{for}P\phantom{\rule{0ex}{0ex}}{P}_{v}\left({P}_{v}\right)=\left(\text{constant}\right){e}^{\frac{{P}_{v}V}{NkT}}\dots \dots \left(2\right)\phantom{\rule{0ex}{0ex}}Dividetheequation\left(1\right)withtheequation\left(2\right),\phantom{\rule{0ex}{0ex}}\frac{{P}_{v}\left(P\right)}{{P}_{v}\left({P}_{v}\right)}=\frac{\left(\text{constant}\right){e}^{\frac{PV}{NkT}}}{\left(\text{constant}\right){e}^{\frac{PV}{NkT}}}\phantom{\rule{0ex}{0ex}}{P}_{v}\left(P\right)={P}_{v}\left({P}_{v}\right){e}^{\frac{PV}{NkTPkT}}\phantom{\rule{0ex}{0ex}}={P}_{v}\left({P}_{v}\right){e}^{\frac{\left(P-{P}_{v}\right)V}{NkT}}$

As a result, the differential equation solution found in part (a) is ${P}_{v}\left({P}_{v}\right){e}^{\frac{\left(P-{P}_{v}\right)}{NkT}}$

## Part (c) Step 4: Explanation

The vapour pressure of water at a temperature of 25°C is as follows.

$P=0.03bar$

As a result, when the total pressure is 1 bar, the vapour pressure is minimal.

$P=1bar\phantom{\rule{0ex}{0ex}}P=1\mathrm{bar}\left(\frac{1.013×{10}^{5}\mathrm{N}/{\mathrm{m}}^{2}}{1\mathrm{bar}}\right)\phantom{\rule{0ex}{0ex}}P=1.013×{10}^{5}\mathrm{N}/{\mathrm{m}}^{2}$

role="math" localid="1646939469352" $\text{Calculate the exponential value of the equation}{P}_{v}\left({P}_{v}\right){e}^{\frac{\left(P-{P}_{v}\right)V}{NkT}}\text{.}\phantom{\rule{0ex}{0ex}}{e}^{\frac{\left(P-{P}_{2}\right)V}{NkT}}={e}^{\frac{PV}{NkT}}$

The value of Nk is equal to the gas constant for one mole of N, and the volume is

$V=18×{10}^{-6}{\mathrm{m}}^{3}$

Therefore,

${e}^{\frac{PV}{NRT}}={e}^{\frac{PV}{RT}}$

Substituting the values

$=\mathrm{exp}\left(\frac{\left(1.013×{10}^{5}\mathrm{N}/{\mathrm{m}}^{2}\right)\left(18×{10}^{-6}{\mathrm{m}}^{3}\right)}{\left(8.31\mathrm{J}/\mathrm{K}\right)\left(\left(25+273\right)\mathrm{K}\right)}\right)\phantom{\rule{0ex}{0ex}}=1.00074$

## Step 5: Conclusion

According to the above calculation, the vapour pressure is 0.07 percent higher than it would be in the absence of air.

Because the volume of the liquid is substantially smaller than that of the gas, the quantity $\frac{PV}{NkT}$ is always less than 1.

At the critical moment, the ratio is comparable to one.

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