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 calculate the total work ww done by the gas when the expansion is carried out in two steps, we use the formula for work in an isobaric (constant pressure) process: w=−PextΔVw = -P_{\text{ext}} \Delta V
Step-by-Step Calculation:
Step 1: Expansion from 1.00 L to 16.0 L at 1.50 atm ΔV1=16.0 L−1.00 L=15.0 L\Delta V_1 = 16.0\,L – 1.00\,L = 15.0\,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 at 1.00 atm ΔV2=24.0 L−16.0 L=8.0 L\Delta V_2 = 24.0\,L – 16.0\,L = 8.0\,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 (in L·atm): wtotal=w1+w2=−22.5−8.0=−30.5 L\cdotpatmw_{\text{total}} = w_1 + w_2 = -22.5 – 8.0 = -30.5\,\text{L·atm}
To convert to Joules: 1 L\cdotpatm=101.325 J1\,\text{L·atm} = 101.325\,\text{J} w=−30.5 L\cdotpatm×101.325 J/L\cdotpatm=−3090.41 Jw = -30.5\,\text{L·atm} \times 101.325\,\text{J/L·atm} = -3090.41\,\text{J}
Explanation (300+ words):
In thermodynamics, work done by a gas during expansion or compression depends on the external pressure and the change in volume. For a reversible or multi-step process, the total work done is the sum of the work done in each step. This problem involves two sequential expansions under different constant external pressures.
In the first step, the gas expands from 1.00 L to 16.0 L against an external pressure of 1.50 atm. The work done is negative because the gas is doing work on the surroundings, which by convention is taken as energy leaving the system. The volume change is significant (15.0 L), and because the pressure is higher, this step accounts for most of the work.
In the second step, the gas continues to expand from 16.0 L to 24.0 L, but against a lower external pressure of 1.00 atm. This means less work is done per liter of expansion, but since the gas is still expanding, this adds additional (negative) work.
Adding both steps gives a total of −30.5-30.5 L·atm. When converted to Joules using the standard conversion factor 1 L\cdotpatm=101.325 J1\,\text{L·atm} = 101.325\,\text{J}, the result is approximately −3090.41 J-3090.41\,\text{J}.
It’s important to note that the total work is less negative than in a single-step expansion (e.g., from 1.00 L to 24.0 L at 1.00 atm only), because in the first step the gas expands against a higher pressure, meaning more work is required. This demonstrates how the path of a thermodynamic process affects the work done.