# Thermal Equilibrium With Insulated Liquid And Gas Containers

1. Apr 26, 2014

### mchahal22

1. The problem statement, all variables and given/known data
A beaker with a metal bottom is filled with 20g of water at 20∘C. It is brought into good thermal contact with a 4000 cm^3 container holding 0.40mol of a monatomic gas at 10atm pressure. Both containers are well insulated from their surroundings.

What is the gas pressure after a long time has elapsed? You can assume that the containers themselves are nearly massless and do not affect the outcome.

2. Relevant equations
Q=mc(delta T)
P=kA(delta T/distance between objects)
pV=nRT

3. The attempt at a solution
I used the ideal gas law to find the initial temp of the gas to be 1219.013 K or 945.863 degrees Celsius. From there, I do not know how to equate the substances in the two containers to determine an equilibrium point and find the final pressure or if this is even the right approach. I would greatly appreciate any help.

2. Apr 26, 2014

### Staff: Mentor

The heat capacity of the water is 1 cal/(gm C) and the molar heat capacity of the gas is 3R/2 = 3 cal/(mole C). The amount of heat gained by the water is equal to the amount of heat lost by the gas. Their final temperatures are the same.

Chet

3. Apr 26, 2014

### mchahal22

So I set the two amounts of heat equal to each other:

mc deltaT (water) = mc delta T (gas)

.02 kg x 4190 J/kgC x (T-20) = - (T-945.863) x m x c

I'm not really getting what m and c would be for the right side of the equation.

m x c = .4 mol x (3x4190) J/moleC ? Is that the correct conversion?

4. Apr 26, 2014

### Staff: Mentor

Yes. But it would have been easier if you stuck to calories.

20 gm x 1 cal/gmC x (T-20) = - (T-945.863) x .4 mol x 3 cal/moleC