When you open a bottle of beer, you can fleetingly see a bit of vapor in the ope

Question

When you open a bottle of beer, you can fleetingly see a bit of vapor in the opening. Here you are to analyses what happens. The gas in the bottle consists mainly of CO_2, with a little bit of water vapor. We will consider it to be like a gas confined in a cylinder by a piston. The pressure in the piston is 0.3 MPa. At the moment that we open the bottle, the pressure drops instantaneously to 0.1 MPa because the piston moves outwards. The conditions in the two situations are: before opening: T_alpha = 280 K., p_alpha = 0.3 MPa after opening: T_beta unknown, p_beta = 0.1 MPa We arbitrarily consider one mole of gas (the amount is not important). The initial volume is V_alpha, = (RT_alpha)/p_alpha. V_beta is unknown because we do not (yet) know the temperature after expansion. The cylinder expands against the external pressure p_beta; in doing so it does an amount of work on the surroundings: W = -P_beta(V_beta - V_alpha). Calculate the temperature immediately after the expansion. Why do we see the little cloud of water vapour?

Explanation / Answer

This is adiabatic expansion agianst constant external pressure.

Work done in the process, w = -Pex(V2-V1) .........(i)

From first law, dU = q + w

Here, q=0, so w= dU =nCvdT = nCv(T2-T1) = -Pex(V2-V1) ...............(ii)

As external pressure against gas expands is equal to P2 = 0.1 MPa, from equation(ii),for 1 mole of a gas, we have

Cv(T2-T1) = -P2(RT2/P2 - RT1/P1) = R(T1*P2/P1 - T2) ................(iii)

Here taking Cp for water = 75.4Jmol-1K-1, and using Cv = Cp - R, we have Cv = 67.1Jmol-1K-1(approximately)

Given T1 = 280 K, P1 = 0.3 MPa, P2 = 0.1 MPa

Now we can put all these given data in equation (iii) to calculate T2.

Using these data in equation (iii), we have T2 = 260 K(approx)

As gas cools suddenly below its freezing point, hence we see little cloud of water vapour.

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