Thermal Equilibrium With Insulated Liquid And Gas Containers

[mchahal22]

Homework Statement

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.

Homework Equations

Q=mc(delta T)
P=kA(delta T/distance between objects)
pV=nRT

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.

 

[Chestermiller]

Homework Statement

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.

Homework Equations

Q=mc(delta T)
P=kA(delta T/distance between objects)
pV=nRT

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.

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

 

[mchahal22]

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

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?

 

[Chestermiller]

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?

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