A gas is confined to a container with a massless piston at the top. (Part C figure) A massless wire is attached to the piston. When an external pressure of 2.00 atm is applied to the wire, the gas compresses from 4.90 to 2.45 L. When the external pressure is increased to 2.50 atm, the gas further compresses from 2.45 to 1.96 L.
In a separate experiment with the same initial conditions, a pressure of 2.50 atm was applied to the gas, decreasing its volume from 4.90 to 1.96 L in one step.
If the final temperature was the same for both processes, what is the difference between q for the two-step process and q for the one-step process in joules?
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To find the difference between q (heat) for the two-step process and q for the one-step process in joules, we need to use the first law of thermodynamics, which states that ΔU = q - w. Here, ΔU represents the change in internal energy of the system, q represents the heat transferred, and w represents the work done on or by the system.
Since the final temperature is the same for both processes, the internal energy change (ΔU) will be the same. Therefore, we can equate the changes in heat and work for the two processes:
q1 - w1 = q2 - w2
We need to calculate the value of q1 and q2 separately, and then find the difference between them.
For the two-step process:
Initial volume (V1) = 4.90 L
Final volume (V2) = 1.96 L
External pressure (P1) = 2.00 atm
External pressure (P2) = 2.50 atm
To calculate q1, we can use the formula:
q1 = ΔU + w1
Since the gas is compressing, the work done on the gas by the external pressure can be calculated as:
w1 = -P1ΔV
ΔV = V2 - V1 = 1.96 L - 4.90 L = -2.94 L
Substituting the values into the equation, we have:
w1 = -2.00 atm * (-2.94 L) = 5.88 L·atm
To find q1, we need the change in internal energy (ΔU), which can be calculated using the ideal gas law:
ΔU = nCvΔT
where n is the number of moles of gas and Cv is the molar heat capacity at constant volume.
Since the final temperature is the same, ΔT = 0. Therefore, ΔU = 0 for both processes.
Now, let's calculate q2 for the two-step process. We can use the formula:
q2 = ΔU + w2
Since ΔU is zero, we only need to calculate w2:
w2 = -P2ΔV
ΔV = V2 - V1 = 1.96 L - 2.45 L = -0.49 L
Substituting the values into the equation, we have:
w2 = -2.50 atm * (-0.49 L) = 1.225 L·atm
Now, we can find the difference between q1 and q2:
Δq = (q1 - q2) = (0 + 5.88 L·atm) - (0 + 1.225 L·atm) = 4.655 L·atm
To convert L·atm to joules, we know that 1 L·atm = 101.3 J. Therefore, the difference in q (heat) between the two-step process and the one-step process is:
Δq = 4.655 L·atm * 101.3 J/L·atm = 471.3 J