Question

2.7 Bottle of beer When you open a bottle of beer, you can fleetingly see a bit of vapour in the opening. Here you are to analyse what happens. The gas in the bottle consists mainly of COa, with a little bit of water vapour. We will consider it to be like a gas v confined in a cylinder by a piston. The pressure in the piston is o.3 MPa. At the moment that we open the bottle, 兀 Pe Ceasos, Thermodynamics for chemical e Aston University the pressure drops instantaneously to o.1 MPa because the piston moves outwards The conditions in the two situations are: 1. before opening T, -280K, p. 0.3MPa 2, after opening :7, unknown, Po=0.1 MPa We arbitrarily consider one mole of gas (the amount is not important). The initial volume is V. = (RT, ) / Pa . , is unknown because we do not (yet) know the temperature after expansion. The eylinder expands against the external pressure psi in doing so it does an amount of beuise we do not Gyeo) know the temperature after sqpansion. The work on the surroundings: W=-pr(K' ) Caleulate the temperature immediately after the expansion. Why do we see the little cloud of water vapour? Note: Cp can be obtained from the tables and while Cp- Cv+R. R -8.314 J mol K* Sol: Blackboard.

Please do it on a paper so i can see the equations clearly.

The answer : T(beta) = 238.4 K

Answer 1

When the mouth of the beer bottle is opened the liquid inside it is directly open to atmosphere. Now,

the process takes place in 2 stages

1. adiabatic process ( expansion)                          2. Isochoric process ( Volume is constant)

1. Adiabatic process

Assume the initial parameters as T₁,V₁,P₁ and after expansion as T_e,V_e, P_e

Given T₁ = 280 k   P₁ = 0.3 MPa        P_e = 0.1013 MPa ( 1 Atmosphere) as it is opened to air

Using $PV^\gamma = constant$

we get $P^1^-^\gamma T^{\gamma }= constant$

Assuming $\gamma$ = 1.289 for CO₂

$P_{1}^{1-\gamma } T_{1}^{\gamma } = P_{e}^{1-\gamma } T_{e}^{\gamma }$

T_e = 218.87 K

2. Isochoric Process

As the expansion started the beer starts coming out due to the sudden expansion from a small volume to a very large volume and after a small time interval the volume of it is fixed.

$\frac{P_{e}}{T_{e}} =\frac{P_{2}}{T_{2}}$

P₂ = 0.1 MPa

T₂ = 0.1 * 218.87 / 0.1013 = 216.06 K

Generally process 2 will be neglected because the drop in temperature is very small

The little cloud of water vapour is due to the absorption of heat from the surroundings by the water molecules that are present on the surface.