Two equal-volume compartments of a box are joined by a thin wall as shown in the figure. The left compartment is filled with 0.050 mole of helium gas at 403 K, and the right compartment contains 0.025 mole of helium gas at 213 K. The right compartment also has a piston to which a force of 20.0 N is applied.
a) If the wall between the compartments is removed, what will the final temperature (in K) of the system be?
b) How much heat will be transferred from the left compartment to the right compartment?
c) What will be the displacement of the piston due to this transfer or heat?
d) What fraction of the heat will be converted into work?
To solve this problem, we can use the ideal gas law, the equation of state for an ideal gas, which states:
PV = nRT
Where:
P = pressure of the gas
V = volume of the gas
n = number of moles of gas
R = ideal gas constant
T = temperature of the gas
We can also use the First Law of Thermodynamics, which states:
ΔU = Q - W
Where:
ΔU = change in internal energy of the system
Q = heat added to the system
W = work done by the system
Now let's solve the questions step by step:
a) To find the final temperature of the system when the wall between the compartments is removed, we can assume that the total volume of the system remains constant. Since the compartments have equal volumes, the volume remains the same even when the wall is removed.
To find the final temperature, we can use the fact that the total number of moles is conserved in the process. The final number of moles is the sum of the initial number of moles in the two compartments.
n_final = n_left_initial + n_right_initial
= 0.050 mol + 0.025 mol
= 0.075 mol
Using the ideal gas law, we can rearrange it to solve for the final temperature:
T_final = (P_final * V) / (n_final * R)
Since the volume remains the same, we can cancel it out:
T_final = (P_final) / (n_final * R)
Now, let's find the pressure in each compartment:
Pressure in left compartment:
P_left = (n_left * R * T_left_initial) / V_left
Pressure in right compartment:
P_right = (n_right * R * T_right_initial) / V_right
Since the volumes are equal, V_left = V_right = V. Substituting this in the equations:
P_left = (n_left * R * T_left_initial) / V
P_right = (n_right * R * T_right_initial) / V
The total pressure at the final state is the sum of the pressures from both compartments:
P_final = P_left + P_right
Substituting the values:
P_left = (0.050 mol * R * 403 K) / V
P_right = (0.025 mol * R * 213 K) / V
P_final = [(0.050 mol * R * 403 K) + (0.025 mol * R * 213 K)] / V
Now substitute P_final in the equation to find T_final:
T_final = [(0.050 mol * R * 403 K) + (0.025 mol * R * 213 K)] / (0.075 mol * R)
= [0.050 * 403 + 0.025 * 213] K
= 20.15 + 5.325 K
= 25.475 K
Therefore, the final temperature of the system will be approximately 25.475 K.
b) To find the amount of heat transferred from the left compartment to the right compartment, we can use the First Law of Thermodynamics:
ΔU = Q - W
Since we assume the total volume remains constant during the process, the change in internal energy (ΔU) is zero. Thus, the equation simplifies to:
0 = Q - W
In this case, if there is no work done (W = 0), then all the heat transferred is equal to Q.
Hence, the amount of heat transferred from the left compartment to the right compartment is equal to the total change in internal energy, which is zero in this case.
c) Since there is no heat transfer (ΔU = 0) and no work done (W = 0), the displacement of the piston due to the transfer of heat will also be zero.
d) Since there is no work done (W = 0) and no heat transferred (Q = 0), the fraction of heat converted into work is also zero.
Therefore, no heat is converted into work in this process.
Note: It is important to mention that this explanation assumes an ideal gas behavior and does not consider any non-ideal effects or complexities that might exist in a real system.