Find the work done when one mole of a gas is expanded reversibly and isothermally from 5 ATM to 1 ATM at 25°c.
(R = 8.314 Joules per kilogram per mole)
First, to find the work done, we can use the formula for work done during an isothermal reversible expansion:
W = -nRT * ln(V2/V1),
where:
W = work done
n = amount of gas (1 mole)
R = gas constant (8.314 J/mol K)
T = temperature (25°C = 25 + 273.15 = 298.15 K)
V1 = initial volume
V2 = final volume
(V2/V1) = ratio of final volume to initial volume.
We can find V1 and V2 using the ideal gas law:
PV = nRT
Initial state:
P1 = 5 atm, V1 =?
V1 = nRT / P1
Final state:
P2 = 1 atm, V2 =?
V2 = nRT / P2
Since we need V2/V1, we can simply divide these two equations:
(V2/V1) = (nRT / P2) / (nRT / P1)
The nRT terms cancel out:
(V2/V1) = P1/P2
The problem provides the initial and final pressures as 5 atm and 1 atm, respectively:
(V2/V1) = 5/1 = 5
Now, we can substitute the values into the equation for work done:
W = -nRT * ln(V2/V1)
W = -1 mol * 8.314 J/mol K * 298.15 K * ln(5)
Calculate the natural log of 5:
ln(5) ≈ 1.61
Now, multiply everything together:
W ≈ -1 * 8.314 * 298.15 * 1.61
W ≈ -3978.901 J
The work done during the isothermal reversible expansion is approximately -3978.9 Joules (to one decimal place).
Since the work done is negative, this means the work is done by the gas on its surroundings during the expansion process.