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Q 5.41

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An Introduction to Thermal Physics
Found in: Page 176
An Introduction to Thermal Physics

An Introduction to Thermal Physics

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

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Short Answer

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: dμl=dμg 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

Pv(P)-PvPv=eP-PvV/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.

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Step by Step Solution

Part (a) Step 1: Explanation

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

μg=μ°+kTlnPv

Where

μ° is constant

k is Boltzmann's constant

T is temperature

P is the partial pressure

Differentiating the equation with respect to P

ddPμg=ddPμ°+kTlnPv=0+kT1PvdPvdPdμgdP=kTPvdPvdP

The chemical potential for liquid is

μ1=GN

Where G is Gibbs free energy

Differentiating the equation with respect to pressure P

ddPμl=ddPGNdμldP=1NdGdP

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

dGdP=V

Substitute V for dGdP

dμldP=1N(V)=VN

Step 2: Calculations

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

dμl=dμgdμldP=dμgdP Substitute values of dμ1dP and dμgdP. VN=kTPvdPvdPdPvdP=VNkTPv

Hence, the differential equation is VNkTPv

Part (b) Step 3: Explanation

The solution for the differential equation dPvdP=VNkTPv is in exponential form. Pv(P)=( constant )ePVNRT.(1) Substitute Pv for PPvPv=( constant )ePvVNkT(2)Divide the equation (1) with the equation (2),Pv(P)PvPv=( constant )ePVNkT( constant )ePVNkTPv(P)=PvPvePVNkT PkT=PvPveP-PvVNkT

As a result, the differential equation solution found in part (a) is PvPveP-PvNkT

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= 1 barP=1bar1.013×105 N/m21barP=1.013×105 N/m2

role="math" localid="1646939469352" Calculate the exponential value of the equation PvPveP-PvVNkT. eP-P2VNkT=ePVNkT

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

V=18×10-6 m3

Therefore,

ePVNRT=ePVRT

Substituting the values

=exp1.013×105 N/m218×10-6 m3(8.31 J/K)((25+273)K)=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 PVNkT is always less than 1.

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

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