One mole of an ideal gas undergoes an isothermal reversible expansion, doing 1 kJ of work and doubling the volume. What is its final temperature? How much heat was added to the gas?
To find the final temperature of the gas undergoing the isothermal reversible expansion, we can use the ideal gas law.
The ideal gas law states that:
PV = nRT
Where:
P is the pressure of the gas
V is the volume of the gas
n is the number of moles of gas
R is the ideal gas constant
T is the temperature of the gas
Since it is an isothermal process, the initial and final temperatures are the same. Let's assume the initial volume is V1 and the final volume is V2 (which is double the initial volume).
According to the ideal gas law, we have:
P1V1 / T1 = P2V2 / T2
Since the process is reversible, the pressure remains constant throughout the expansion, so P1 = P2 = P.
Let's substitute in the given values:
P1 * V1 / T1 = P * V2 / T2
Using the fact that V2 = 2 * V1, we can simplify the equation to:
P * V1 / T1 = P * (2 * V1) / T2
Cancelling out P and V1, we are left with:
1 / T1 = 2 / T2
Simplifying further, we get:
T2 = T1 / 2
Therefore, the final temperature (T2) is equal to half the initial temperature (T1).
To determine the amount of heat added to the gas, we can use the first law of thermodynamics:
ΔU = Q - W
Where:
ΔU is the change in internal energy of the gas
Q is the amount of heat added to the gas
W is the work done by the gas
In this case, the internal energy change is zero because it is an isothermal process. So, we have:
0 = Q - W
Since the work done by the gas is given as 1 kJ, we have:
0 = Q - 1 kJ
Solving for Q, we find that the amount of heat added to the gas is 1 kJ.
So, the final temperature is half the initial temperature, and the heat added to the gas is 1 kJ.