A sample of an ideal gas at 15.0 atm and 10.0 L is allowed to expand against a constant external pressure of 2.00 atm at a constant temperature. Calculate the work in units of kJ for the gas expansion. (Hint: Boyle's law applies)
Please explain thoroughly! thank you!
To calculate the work done during the gas expansion, we need to use Boyle's law, which states that the pressure and volume of an ideal gas are inversely proportional at constant temperature.
First, we need to determine the initial and final volumes of the gas.
The initial volume is given as 10.0 L.
To find the final volume, we apply Boyle's law:
(P1)(V1) = (P2)(V2)
Where:
P1 = initial pressure = 15.0 atm
V1 = initial volume = 10.0 L
P2 = final pressure = 2.00 atm
Rearranging the equation to solve for V2, we have:
V2 = (P1)(V1) / P2
Plugging in the values:
V2 = (15.0 atm)(10.0 L) / 2.00 atm
V2 = 75.0 L
Now we have both the initial and final volumes. The work done in an isobaric (constant pressure) process is given by:
Work = (P2 - P1) * (V2 - V1)
Where:
P1 = initial pressure = 15.0 atm
P2 = final pressure = 2.00 atm
V1 = initial volume = 10.0 L
V2 = final volume = 75.0 L
Plugging in the values:
Work = (2.00 atm - 15.0 atm) * (75.0 L - 10.0 L)
Work = (-13.0 atm) * (65.0 L)
Work = -845.0 atm⋅L
The work is given in units of atm⋅L, but we need to convert it to kJ.
To convert from atm⋅L to joules, we use the conversion factor:
1 atm⋅L = 101.3 J
To convert from joules to kilojoules, divide by 1000:
1 J = 0.001 kJ
Now, let's convert the units:
Work in joules = -845.0 atm⋅L * 101.3 J/atm⋅L
Work in joules = -85551.5 J
Work in kilojoules = -85551.5 J * (0.001 kJ/1 J)
Work in kilojoules ≈ -85.55 kJ
Therefore, the work done during the gas expansion is approximately -85.55 kJ. The negative sign indicates that work is done on the gas rather than by the gas.