Find the work done when one mole of the gas is expanded reversibly and isothermally from 5 atm to 1 atm at 25 degree Celsius
We can use the formula for work done during an isothermal reversible expansion of an ideal gas:
W = nRT ln(Vf / Vi)
Where n is the number of moles (given as 1 mole), R is the ideal gas constant (8.314 J/mol · K), T is the temperature (given as 25 °C, which is 298 K), and Vf and Vi are the final and initial volumes, respectively.
However, we need to find the Vf and Vi first. We can use the ideal gas law:
PV = nRT
Vi = nRT / Pi
Vf = nRT / Pf
Plugging in the values, and plugging in 5 atm and 1 atm as initial and final pressure, respectively,
Vi = (1 mol * 8.314 J/mol · K * 298 K) / (5 atm)
Vf = (1 mol * 8.314 J/mol · K * 298 K) / (1 atm)
We need to make sure our units are consistent. I'll convert the gas constant R from J/mol·K to L·atm/mol·K. The conversion is 1 L·atm = 101.325 J:
R = 8.314 J/mol·K * (1 L·atm / 101.325 J) = 0.0821 L·atm/mol·K
Now, we can calculate Vi and Vf:
Vi = (1 mol * 0.0821 L·atm/mol·K * 298 K) / (5 atm) = 4.923 L
Vf = (1 mol * 0.0821 L·atm/mol·K * 298 K) / (1 atm) = 24.615 L
Now, we can plug these values into the formula for the work done:
W = (1 mol * 0.0821 L·atm/mol·K * 298 K) * ln(24.615 L / 4.923 L)
W = (24.37 L·atm) * ln(5)
W ≈ -58.13 L·atm
To convert the work back to joules, multiply by the conversion factor:
W = -58.13 L·atm * (101.325 J / 1 L·atm)
W ≈ -5886 J
The work done when one mole of the gas is expanded reversibly and isothermally from 5 atm to 1 atm at 25 °C is approximately -5886 J. Since the work is negative, it indicates that work is being done by the gas on the surroundings during the expansion.