Take 3.00 mol of an ideal gas at Pi=25.0 bar, Vi=4.50L, and expand it isothermally. Calculate the work done in two different processes.
a. Reversible expansion at Pf=4.50 bar
b. Irreversible expansion at Pext=4.50 bar
To calculate the work done in each process, we need to use the formula for work done in an isothermal expansion:
W = -nRT ln(Vf/Vi)
where W is the work done, n is the number of moles, R is the ideal gas constant, T is the temperature in Kelvin, Vf is the final volume, and Vi is the initial volume.
First, let's find the temperature (T) of the gas in Kelvin. We can use the ideal gas equation:
PV = nRT
Rearranging the equation gives us:
T = PV / (nR)
Plugging in the given values, we have:
T = (25.0 bar * 4.50 L) / (3.00 mol * 0.0821 L·atm/(mol·K))
T = 38.19 K
a. Reversible Expansion:
For a reversible process, we can assume that the expansion is happening slowly, allowing the system to be close to equilibrium at each step. In this case, the final pressure is given as Pf = 4.50 bar.
Using the equation for work done, we have:
W = -nRT ln(Vf/Vi)
Plugging in the values, we get:
W = -(3.00 mol) * (0.0821 L·atm/(mol·K)) * (38.19 K) * ln(4.50 L / 4.50 L)
Since ln(1) = 0, the work done for the reversible expansion (a) is zero.
b. Irreversible Expansion:
For an irreversible process, the pressure of the system is not necessarily equal to the external pressure. In this case, the external pressure is given as Pext = 4.50 bar.
Using the same formula for work done, we have:
W = -nRT ln(Vf/Vi)
Plugging in the values, we get:
W = -(3.00 mol) * (0.0821 L·atm/(mol·K)) * (38.19 K) * ln(4.50 L / 4.50 L)
Since ln(1) = 0, the work done for the irreversible expansion (b) is also zero.
Therefore, for both the reversible and irreversible expansion, the work done is zero.