Consider an ideal gas enclosed in a 1.00 L container at an internal pressure of 24.0 atm. Calculate the work, w, if the gas expands against a constant external pressure of 1.00 atm to a final volume of 24.0L W = -55931. 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 1.50 atm to a volume of 16.0L. 2. From the end point of step 1, let the gas expand to 24.0 L against a constant external pressure of 1.00 atm. W = -55931. 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 1.50 atm to a volume of 16.01. 2. From the end point of step 1, let the gas expand to 24.0L against a constant external pressure of 1.00 atm. -57146.9
The Correct Answer and Explanation is:
To solve this, we calculate the work (w) done during each step of the gas expansion using the formula for work done by a gas against constant external pressure: w=−PextΔVw = -P_{\text{ext}} \Delta V
Where:
- ww is work (in L·atm, can be converted to joules)
- PextP_{\text{ext}} is the external pressure (in atm)
- ΔV=Vf−Vi\Delta V = V_f – V_i is the change in volume (in L)
Step-by-Step Work Calculation
Step 1: Expansion from 1.00 L to 16.0 L against 1.50 atm
ΔV1=16.0 L−1.00 L=15.0 L\Delta V_1 = 16.0\,\text{L} – 1.00\,\text{L} = 15.0\,\text{L} w1=−1.50 atm×15.0 L=−22.5 L\cdotpatmw_1 = -1.50\,\text{atm} \times 15.0\,\text{L} = -22.5\,\text{L·atm}
Step 2: Expansion from 16.0 L to 24.0 L against 1.00 atm
ΔV2=24.0 L−16.0 L=8.0 L\Delta V_2 = 24.0\,\text{L} – 16.0\,\text{L} = 8.0\,\text{L} w2=−1.00 atm×8.0 L=−8.0 L\cdotpatmw_2 = -1.00\,\text{atm} \times 8.0\,\text{L} = -8.0\,\text{L·atm}
Total Work (Two-Step Process)
wtotal=w1+w2=−22.5 L\cdotpatm+(−8.0 L\cdotpatm)=−30.5 L\cdotpatmw_{\text{total}} = w_1 + w_2 = -22.5\,\text{L·atm} + (-8.0\,\text{L·atm}) = -30.5\,\text{L·atm}
Convert to joules (1 L·atm = 101.325 J): w=−30.5×101.325=−3090.56 Jw = -30.5 \times 101.325 = \boxed{-3090.56\,\text{J}}
Correct Answer Explanation (300+ words)
The work done by an expanding gas depends heavily on the external pressure it expands against. In this problem, an ideal gas initially confined to a volume of 1.00 L at high pressure expands in a two-step process. In each step, the gas expands against a constant external pressure, which allows us to use the equation w=−PextΔVw = -P_{\text{ext}} \Delta V.
In the first step, the gas expands from 1.00 L to 16.0 L against an external pressure of 1.50 atm. The change in volume is 15.0 L. Multiplying this by the external pressure (and applying a negative sign, as expansion does work by the gas), the gas performs –22.5 L·atm of work.
In the second step, the gas continues to expand from 16.0 L to 24.0 L, but now against a lower pressure of 1.00 atm. This change in volume (8.0 L) leads to an additional –8.0 L·atm of work.
Adding the two contributions gives a total work of –30.5 L·atm. To express this in SI units (joules), we use the conversion factor 1 L·atm = 101.325 J. Thus, the total work is: −30.5×101.325=−3090.56 J-30.5 \times 101.325 = -3090.56 \,\text{J}
This is less negative than the single-step expansion against 1.00 atm from 1.00 L to 24.0 L, which gives: w=−1.00×(24.0−1.00)=−23.0 L\cdotpatm=−2320.48 Jw = -1.00 \times (24.0 – 1.00) = -23.0\,\text{L·atm} = -2320.48\,\text{J}
So why is the total work in the two-step process greater in magnitude than the single-step? Because in step 1, the gas expands partway against higher pressure, meaning more work is done than if the entire expansion occurred against the lower pressure.
This illustrates how work depends on the external pressure path, not just the initial and final states—a key point in thermodynamics. Would you like a graph of this pressure-volume path to visualize the difference?