What is the value of w (work) when 2.6 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.

To find the value of work (w) done in each step of the expansion, you can use the formula:

w = -Pext * ΔV

where Pext is the external pressure and ΔV is the change in volume.

Step 1:
In the first step, the volume change is from 25 liters to 50 liters, and the external pressure is 1 atm. Plugging these values into the formula:

w1 = -1 atm * (50 L - 25 L)

Simplifying, we get:

w1 = -1 atm * 25 L
w1 = -25 atm L

Since the answer needs to be in calories, we need to convert from atm L to calories. As 1 atm L = 101.3 J, and 1 calorie = 4.184 J, we can convert as follows:

w1 = -25 atm L * (101.3 J / 1 atm L) * (1 cal / 4.184 J)
w1 ≈ -605 cal

Step 2:
In the second step, the volume change is from 50 liters to 100 liters, and the external pressure is 0.1 atm. Applying the formula:

w2 = -0.1 atm * (100 L - 50 L)

Simplifying, we get:

w2 = -0.1 atm * 50 L
w2 = -5 atm L

Converting to calories:

w2 = -5 atm L * (101.3 J / 1 atm L) * (1 cal / 4.184 J)
w2 ≈ -120 cal

The total work done (wtot) is the sum of w1 and w2:

wtot = w1 + w2
wtot ≈ -605 cal + -120 cal
wtot ≈ -725 cal

Therefore, the value of work (w) when 2.6 moles of H2 undergoes the specified expansion is approximately -725 cal.