Question

consider an ideal gas enclosed in a 1.00 l container at an internal pressure of 20.0 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.

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 2.00 atm to a volume of 10.0 l.
2. from the end point of step 1, let the gas expand to 20.0 l against a constant external pressure of 1.00 atm.

Explanation:

Step1: Calculate work for first - step expansion

The formula for work done in expansion against a constant external pressure is \(w=-P_{ext}\Delta V\).
For the first - step, \(P_{ext1} = 2.00\ atm\), \(V_{1}=1.00\ L\), \(V_{2} = 10.0\ L\).
\(\Delta V_1=V_{2}-V_{1}=10.0 - 1.00=9.00\ L\)
\(w_1=-P_{ext1}\Delta V_1=-2.00\ atm\times9.00\ L=- 18.0\ L\cdot atm\)

Step2: Calculate work for second - step expansion

For the second - step, \(P_{ext2}=1.00\ atm\), \(V_{2} = 10.0\ L\), \(V_{3}=20.0\ L\)
\(\Delta V_2=V_{3}-V_{2}=20.0 - 10.0 = 10.0\ L\)
\(w_2=-P_{ext2}\Delta V_2=-1.00\ atm\times10.0\ L=-10.0\ L\cdot atm\)

Step3: Convert units and find total work

We know that \(1\ L\cdot atm = 101.325\ J\)
\(w_1=-18.0\ L\cdot atm\times101.325\ J/L\cdot atm=-1823.85\ J\)
\(w_2=-10.0\ L\cdot atm\times101.325\ J/L\cdot atm=-1013.25\ J\)
\(w = w_1 + w_2=-(1823.85 + 1013.25)\ J=-2837.1\ J\)

Answer:

\(-2837.1\ J\)