What is the value of w (work) when 1.3 moles

of H2 expands from 25 liters to 50 liters
against a constant external pressure of 1 atm,
then expands further from 50 to 100 liters
against a constant external pressure of 0.1
atm? Answer in calories.
Answer in units of cal
I know to use the P*deltaV formula but whatever I'm doing isn't working

I would try to do each step separately.

1*25 = -25 L*atm
0.1*50 = -5 L*atm
Total is -30 L*atm and convert L*atm to calories. I don't know the conversion factor BUT L*atm to joules is x 101.325 and joules to calories is divide by 4.184.

To calculate the work done (w) in this scenario, you can use the formula w = P * ΔV, where P is the pressure and ΔV is the change in volume.

First, let's calculate the work done during the first expansion from 25 liters to 50 liters against a constant external pressure of 1 atm.

Step 1: Convert the pressure from atm to calories.
1 atm = 1.01325 × 10^3 g⋅cm^2⋅s^-2.
Since calories are defined as the amount of heat required to raise the temperature of 1 gram of water by 1 degree Celsius, 1 atm is equal to 1.01325 × 10^3 cal.

Step 2: Calculate the change in volume.
ΔV = V_final - V_initial = 50 L - 25 L = 25 L.

Step 3: Calculate the work done.
w = P * ΔV = 1 atm * 25 L = 25 atm⋅L.

Step 4: Convert the work done to calories.
To convert from atm⋅L to cal, you need to multiply by the conversion factor: 1 atm⋅L = 101.325 J = 24.216 cal.

Therefore, the work done in the first expansion is:
w1 = 25 atm⋅L * 24.216 cal = 605.4 cal.

Now, let's calculate the work done during the second expansion from 50 liters to 100 liters against a constant external pressure of 0.1 atm.

Step 1: Convert the pressure from atm to calories.
0.1 atm = 0.101325 × 10^3 cal.

Step 2: Calculate the change in volume.
ΔV = V_final - V_initial = 100 L - 50 L = 50 L.

Step 3: Calculate the work done.
w2 = P * ΔV = 0.1 atm * 50 L = 5 atm⋅L.

Step 4: Convert the work done to calories.
5 atm⋅L * 24.216 cal = 121.08 cal.

Now, to find the total work done, you need to sum up the work done in the first and second expansions:

Total work done = w1 + w2 = 605.4 cal + 121.08 cal = 726.48 cal.

Therefore, the value of w (work) is 726.48 calories in this case.