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.
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)
1 answer