Heat Transfer Final Temperature Problem

[breez]

There are two insulated compartments separated by a thin wall. The left side contains .060 mol of He at a initial temperature of 600 k and the right side contains .030 mol He at the initial temperature of 300 K. The compartment on the right is attached to a vertical cylinder, above which the air pressure is 1.0 atm. A 10-cm-diameter, 2.0 kg piston can slide without friction up and down the cylinder. The volumes of the compartments are not known, but the cylinder diameter is known.

a. What is the final temperature?
b. How much heat is transferred from the left side to the right side?
c. How high is the piston lifted due to the heat transfer?
d. What fraction of the heat is converted to work?

I believe Part b follows directly from the relation between absolute temperature and KE (just find the difference in temperatures between initial and equilibrium states). However, I cannot figure out part a. Any tips/leads would be helpful.

 

[Stovebolt]

There are two insulated compartments separated by a thin wall. The left side contains .060 mol of He at a initial temperature of 600 k and the right side contains .030 mol He at the initial temperature of 300 K. The compartment on the right is attached to a vertical cylinder, above which the air pressure is 1.0 atm. A 10-cm-diameter, 2.0 kg piston can slide without friction up and down the cylinder. The volumes of the compartments are not known, but the cylinder diameter is known.

a. What is the final temperature?
b. How much heat is transferred from the left side to the right side?
c. How high is the piston lifted due to the heat transfer?
d. What fraction of the heat is converted to work?

I believe Part b follows directly from the relation between absolute temperature and KE (just find the difference in temperatures between initial and equilibrium states). However, I cannot figure out part a. Any tips/leads would be helpful.

Given that you're only dealing with Helium, the final temperature of the entire system will be equal to the average temperature per mole.

You're on the right track for part b. The energy transferred will be directly proportional to the change in temperature of one side.

One thing to note - while the volume of the compartments are not given, there is enough information to calculate the initial volume of the right compartment. This should help with part c.

 

[thomate1]

a. To determine the final temperature, we can use the principle of conservation of energy. Since the compartments are insulated, we can assume that there is no heat exchange with the surroundings. Therefore, the total internal energy of the system (i.e. the combined He in both compartments) must remain constant. We can express this as:

Ufinal = Uinitial

The internal energy of a gas is given by the formula U = (3/2) nRT, where n is the number of moles, R is the gas constant, and T is the temperature in Kelvin. Using this formula for both compartments, we can set up the following equation:

(3/2) (0.060 mol) (R) (Tfinal) + (3/2) (0.030 mol) (R) (Tfinal) = (3/2) (0.060 mol) (R) (600 K) + (3/2) (0.030 mol) (R) (300 K)

Solving for Tfinal, we get:

Tfinal = 450 K

Therefore, the final temperature is 450 Kelvin.

b. To calculate the heat transferred from the left side to the right side, we can use the formula Q = nCΔT, where Q is the amount of heat transferred, n is the number of moles, C is the molar specific heat capacity, and ΔT is the change in temperature. For this problem, we can assume that the molar specific heat capacity of He is constant and equal to 20.8 J/mol K. Therefore, we can calculate the heat transferred as:

Q = (0.060 mol) (20.8 J/mol K) (600 K - 450 K) = 624 J

c. To determine how high the piston is lifted due to the heat transfer, we can use the formula W = PΔV, where W is the work done, P is the pressure, and ΔV is the change in volume. Since the piston is attached to the compartment on the right, we can assume that the volume of the right compartment increases by the same amount that the volume of the left compartment decreases. Therefore, we can calculate the change in volume as:

ΔV = (0.060 mol) (8.314 J/mol K) (600 K - 450 K) / (1.0 atm) = 0.747