Path A
A sample of gas in a cylinder of volume 3.96 L at 327 K and 3.28 atm expands to 7.34 L by two different pathways. Path A is an isothermal, reversible expansion. Calculate the work for Path A.
Answer in units of J.
Path B
Path B has two steps. In the first step, the gas is cooled at constant volume (3.96 L) until its pressure drops to 1.77 atm. In the second step, the gas is heated and allowed to expand against a constant external pressure of 1.77 atm until the final volume is 7.34 L.
Calculate the work for Path B.
Answer in units of J.
Help me please ! Do it as detail as you can, if not you can just leave me with the answer... I will try to figure out how you did it !
Sure! Let's start with Path A, which is an isothermal, reversible expansion. In this case, we can use the equation for the work done during an isothermal process:
Work = -nRT ln(Vf/Vi)
where n is the number of moles of gas, R is the ideal gas constant, T is the temperature in Kelvin, Vi is the initial volume, and Vf is the final volume.
To calculate the work for Path A, we need to know the number of moles of gas. However, since it is not provided in the question, we can use the ideal gas law to find it. The ideal gas law equation is:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
Now, we can rearrange the ideal gas law equation to solve for n:
n = PV / RT
Next, we can substitute this value of n into the work equation and calculate the work. Let's start with the given values:
Initial volume (Vi) = 3.96 L
Final volume (Vf) = 7.34 L
Temperature (T) = 327 K
Now, we need to determine the number of moles (n) using the ideal gas law equation. The given pressure (P) is 3.28 atm. We need to convert it to units of Pascals (Pa) for consistency with the gas constant (R) in SI units.
1 atm = 101325 Pa
3.28 atm = 3.28 * 101325 Pa
Substituting the values into the equation, we have:
n = (3.28 * 101325 Pa * 3.96 L) / (0.08206 L atm / (mol K) * 327 K)
Now, we can calculate 'n':
n = 0.5148 mol
Substituting this value of 'n' into the work equation, we have:
Work = - (0.5148 mol) * (0.08206 L atm / (mol K)) * (327 K) * ln(7.34 L / 3.96 L)
Calculating these values:
Work = - 107.74 J
So, the work for Path A is -107.74 J.
Now, let's move on to Path B, which consists of two steps:
1. In the first step, the gas is cooled at constant volume until its pressure drops to 1.77 atm. Since the volume remains constant, no work is done: Work_stage1 = 0 J.
2. In the second step, the gas is heated and allowed to expand against a constant external pressure of 1.77 atm until the final volume is 7.34 L.
For this step, we can use the equation for the work done during a constant external pressure expansion:
Work = -Pext(Vf - Vi)
where Pext is the external pressure, Vf is the final volume, and Vi is the initial volume.
Given values for Path B:
Initial volume (Vi) = 3.96 L
Final volume (Vf) = 7.34 L
External pressure (Pext) = 1.77 atm
Let's substitute these values into the equation:
Work_stage2 = -(1.77 atm) * (7.34 L - 3.96 L)
Calculating this:
Work_stage2 = -6.7371 J
Now, we can calculate the total work for Path B by adding the work from both stages:
Work_B = Work_stage1 + Work_stage2
Work_B = 0 J + (-6.7371 J)
Work_B = -6.7371 J
So, the work for Path B is -6.7371 J.
I hope this explanation helps you understand how to calculate the work for both Path A and Path B. Let me know if you have any further questions!
To calculate the work for Path A, we can use the formula:
Work = -nRT ln(V2/V1)
Where:
- Work is the work done on or by the gas
- n is the number of moles of the gas
- R is the ideal gas constant (8.314 J/(mol·K))
- T is the temperature in Kelvin
- V2 is the final volume of the gas
- V1 is the initial volume of the gas
Given values:
- V1 = 3.96 L
- T = 327 K
- V2 = 7.34 L
- P = 3.28 atm
First, we need to calculate the number of moles (n) using the ideal gas equation:
PV = nRT
n = PV / RT
n = (3.28 atm * 3.96 L) / (0.0821 L·atm/(mol·K) * 327 K)
n ≈ 0.46 mol
Now we can calculate the work for Path A:
Work = -nRT ln(V2/V1)
Work = -(0.46 mol * 0.0821 L·atm/(mol·K) * 327 K) ln(7.34 L / 3.96 L)
Work ≈ -94 J
Therefore, the work for Path A is approximately -94 J.
For Path B, we need to break it down into two steps and calculate the work for each step separately.
Step 1: Cooling at constant volume
Since the volume is constant, no work is done in this step.
Step 2: Expansion against a constant external pressure
In this step, the work can be calculated using the formula:
Work = -Pext ΔV
Where:
- Work is the work done on or by the gas
- Pext is the external pressure
- ΔV is the change in volume
Given values:
- Pext = 1.77 atm
- ΔV = V2 - V1
ΔV = 7.34 L - 3.96 L
ΔV = 3.38 L
Now we can calculate the work for Step 2:
Work = -(1.77 atm)(3.38 L)
Work ≈ -6 J
Therefore, the work for Path B is approximately -6 J.