Consider an ideal gas enclosed in a 1.00 L container at an internal pressure of

Question

Consider an ideal gas enclosed 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 15.0 L. Number w = 15 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 3.00 L. 2. From there, let the gas expand to 15.0 L against a constant external pressure of 1.00 atm. Number w = J

Explanation / Answer

Solution :-

Formula to calculate the work done at constant pressure is as follows

W= - P delta V

Delta V is the change in the volume

Lets first calculate the work done when gas expands from 1.00 L to 15.0 L at 1.00 atm

W= - 1.00 atm * (15.0 L – 1.00L)

W= -14 L atm

Now lets convert L atm to Joules

1 L atm = 101.325 J

-14 L atm * 101.325 J / 1 L atm = -1488 J

Therefore work done = -1488 J

Part 2 calculating work done in 2 steps

Step 1 at constant pressure 5 atm from 1 L to 3.00 L

W= -5 atm * (3.00 L – 1.00 L)

W= -10 L atm

Converting to Joules

-10 L atm * 101.325 J / 1 L atm = -1013 J

Now calculating at 1 .00 atm from 3.00 L to 15.0 L

W= -1.00 atm * (15.0 L – 3.00 L)

W= -12 L atm

Converting to Joules

12 L atm * 101.325 J / 1 L atm = -1216 Joules

Therefore total work done = -1013 J + (-1216 J) = -2229 J

So total work done = -2229 J

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---