Answer
When the valve is opened and the gas expands to fill the entire system, it does work on the surroundings by pushing against the vacuum in the second cylinder. However, not all of the energy that the gas had initially is available to do work. Some of the energy is lost due to irreversibilities, such as frictional losses and heat transfer to the surroundings. This lost energy is the energy that becomes unavailable to do work.
To find the expression for the unavailable energy, we can start by considering the work done by the gas as it expands. Since the process is isothermal, the work done can be calculated using:
W = -nRT ln(V2/V1)
where W is the work done, n is the number of moles of gas, R is the gas constant, T is the temperature, V1 is the initial volume of the gas in the full cylinder, and V2 is the final volume of the gas in the combined system.
Since the second cylinder was pumped down to vacuum, its initial volume is negligible and we can assume that V2 is equal to the total volume of the two cylinders. Therefore, we have:
V2 = V1 + V0
where V0 is the volume of the second cylinder (which is assumed to be negligible).
Substituting this into the equation for the work done, we get:
W = -nRT ln[(V1 + V0)/V1]
Next, we need to calculate the change in internal energy of the gas during the process. Since the process is isothermal, the change in internal energy is equal to zero.
ΔU = 0
Therefore, the lost energy is equal to the negative of the work done:
ΔW_lost = -W = nRT ln[(V1 + V0)/V1]
This expression gives the energy that becomes unavailable to do work due to irreversibilities during the expansion process. Note that it depends on the initial volume of the gas in the full cylinder and the volume of the second cylinder, but not on the final pressure of the gas.
Work Step by Step
When the valve is opened and the gas expands to fill the entire system, it does work on the surroundings by pushing against the vacuum in the second cylinder. However, not all of the energy that the gas had initially is available to do work. Some of the energy is lost due to irreversibilities, such as frictional losses and heat transfer to the surroundings. This lost energy is the energy that becomes unavailable to do work.
To find the expression for the unavailable energy, we can start by considering the work done by the gas as it expands. Since the process is isothermal, the work done can be calculated using:
W = -nRT ln(V2/V1)
where W is the work done, n is the number of moles of gas, R is the gas constant, T is the temperature, V1 is the initial volume of the gas in the full cylinder, and V2 is the final volume of the gas in the combined system.
Since the second cylinder was pumped down to vacuum, its initial volume is negligible and we can assume that V2 is equal to the total volume of the two cylinders. Therefore, we have:
V2 = V1 + V0
where V0 is the volume of the second cylinder (which is assumed to be negligible).
Substituting this into the equation for the work done, we get:
W = -nRT ln[(V1 + V0)/V1]
Next, we need to calculate the change in internal energy of the gas during the process. Since the process is isothermal, the change in internal energy is equal to zero.
ΔU = 0
Therefore, the lost energy is equal to the negative of the work done:
ΔW_lost = -W = nRT ln[(V1 + V0)/V1]
This expression gives the energy that becomes unavailable to do work due to irreversibilities during the expansion process. Note that it depends on the initial volume of the gas in the full cylinder and the volume of the second cylinder, but not on the final pressure of the gas.
