Consider an ideal gas encloesd in a 1.00 L container at an internal pressure of 10.0 atm. Calculate the work, w, if the gas expands against a constant external pressure of 1.00 atm to a final volume of 20.0 L.
w = ____ J
now calculate the work done if this process is carried out in two steps.
1] first let, the gas expand against a constant external pressure of 5.00 atm to a volume of 4.00 L.
2] From there, let the gas expand to 20.0 L against a constant external pressure of 1.00 atm.
w = _____ J
please help!
sali
To calculate the work done by an ideal gas expanding or compressing, we can use the formula:
w = -Pext * ΔV
where:
w is the work done
Pext is the external pressure
ΔV is the change in volume
For the first part of the question, where the gas expands against a constant external pressure of 1.00 atm to a final volume of 20.0 L:
ΔV = Vf - Vi
= 20.0 L - 1.00 L
= 19.0 L
Therefore, the work done (w) can be calculated as:
w = -1.00 atm * 19.0 L
= -19.0 atm L
Note: The unit of pressure in work calculations is atm, and the unit of volume is L. The negative sign indicates that work is being done on the gas.
For the second part of the question, where the process is carried out in two steps:
Step 1: The gas expands against a constant external pressure of 5.00 atm to a volume of 4.00 L.
ΔV1 = V1f - V1i
= 4.00 L - 1.00 L
= 3.00 L
w1 = -5.00 atm * 3.00 L
= -15.0 atm L
Step 2: From there, the gas expands to 20.0 L against a constant external pressure of 1.00 atm.
ΔV2 = V2f - V2i
= 20.0 L - 4.00 L
= 16.0 L
w2 = -1.00 atm * 16.0 L
= -16.0 atm L
To calculate the total work done (w) for the two-step process, we sum up the individual works:
w = w1 + w2
= (-15.0 atm L) + (-16.0 atm L)
= -31.0 atm L
Therefore, the total work done is -31.0 atm L.