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
Solve for p*delta V at one P and add that for the second expansion.l
To determine the work done (w) in this scenario, we need to use the formula:
w = -PΔV
where w is the work done, P is the pressure, and ΔV is the change in volume.
First, let's calculate the work done for the first expansion from 25 liters to 50 liters against a constant external pressure of 1 atm.
ΔV = 50 L - 25 L = 25 L
Since the pressure is constant at 1 atm, we can substitute these values into the formula:
w1 = -1 atm * 25 L
Now, let's calculate the work done for the second expansion from 50 liters to 100 liters against a constant external pressure of 0.1 atm.
ΔV = 100 L - 50 L = 50 L
Substituting these values into the formula:
w2 = -0.1 atm * 50 L
To find the total work done (w), we add up w1 and w2:
w = w1 + w2
Now, let's calculate the result using the given values:
w1 = -1 atm * 25 L = -25 atm·L
w2 = -0.1 atm * 50 L = -5 atm·L
w = w1 + w2
w = (-25 atm·L) + (-5 atm·L)
w = -30 atm·L
We need to convert the units from atm·L to cal. 1 atm·L is equal to 101.325 J, and 1 cal is equal to 4.184 J. Therefore:
w = -30 atm·L * 101.325 J / 4.184 J/cal
w ≈ -724.96 cal
Now that the work is negative, it signifies that work was done by the system on the surroundings.
Therefore, 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 and then to 100 liters against a constant external pressure of 0.1 atm is approximately -724.96 calories (cal).