Question

consider an ideal gas enclosed in a 1.00 l container at an internal pressure of 14.0 atm. calculate the work, w, if the gas expands against a constant external pressure of 1.00 atm to a final volume of 14.0 l. w = 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 2.80 atm to a volume of 5.00 l. 2. from the end point of step 1, let the gas expand to 14.0 l against a constant external pressure of 1.00 atm.

Explanation:

Step1: Calculate work for single - step expansion

The formula for work done in expansion against a constant external pressure is \(w=-P_{ext}\Delta V\).
Here, \(P_{ext} = 1.00\ atm\), \(V_{i}=1.00\ L\), \(V_{f}=14.0\ L\), and \(\Delta V=V_{f}-V_{i}\).
\(\Delta V=(14.0 - 1.00)\ L=13.0\ L\)
\(w=-P_{ext}\Delta V=- 1.00\ atm\times13.0\ L\)
Using the conversion factor \(1\ L\cdot atm = 101.325\ J\)
\(w=-1.00\times13.0\times101.325\ J=-1317.225\ J\)

Step2: Calculate work for two - step expansion

Step2.1: Work for first step

For the first step, \(P_{ext1}=2.80\ atm\), \(V_{i1} = 1.00\ L\), \(V_{f1}=5.00\ L\)
\(\Delta V_{1}=V_{f1}-V_{i1}=(5.00 - 1.00)\ L = 4.00\ L\)
\(w_{1}=-P_{ext1}\Delta V_{1}=-2.80\ atm\times4.00\ L\)
Using \(1\ L\cdot atm=101.325\ J\)
\(w_{1}=-2.80\times4.00\times101.325\ J=-1134.84\ J\)

Step2.2: Work for second step

For the second step, \(P_{ext2}=1.00\ atm\), \(V_{i2}=5.00\ L\), \(V_{f2}=14.0\ L\)
\(\Delta V_{2}=V_{f2}-V_{i2}=(14.0 - 5.00)\ L = 9.00\ L\)
\(w_{2}=-P_{ext2}\Delta V_{2}=-1.00\ atm\times9.00\ L\)
Using \(1\ L\cdot atm = 101.325\ J\)
\(w_{2}=-1.00\times9.00\times101.325\ J=-911.925\ J\)

Step2.3: Total work for two - step

\(w_{total}=w_{1}+w_{2}\)
\(w_{total}=(-1134.84)+(-911.925)\ J=-2046.765\ J\)

Answer:

• For single - step expansion: \(w=-1317\ J\)
• For two - step expansion: \(w=-2047\ J\)