A sample of gas in a cylinder of volume 3.64 L at 275 K and 2.68 atm expands to 7.04 L by two different pathways. Path A is an isothermal, reversible expansion. Calculate the work for Path A.
Answer in units of J
What I did was basically W=-P(delta V)
Which was -(2.68 atm)(3.4 L)(101.3 J/L*atm) and ended up getting -923.0456 J, but that was not the correct answer. So, how would I do this problem then?
To calculate the work done for an isothermal, reversible expansion (Path A), you need to use the formula:
Work (W) = -nRT ln(Vf/Vi)
Where:
- n is the number of moles of gas
- R is the ideal gas constant (0.0821 L·atm/(mol·K))
- T is the temperature in Kelvin
- Vi is the initial volume in liters
- Vf is the final volume in liters
- ln represents the natural logarithm.
Before we calculate the work, we need to determine the number of moles of gas using the ideal gas equation:
PV = nRT
Rearranging the equation:
n = PV / RT
Substituting the given values:
n = (2.68 atm) (3.64 L) / (0.0821 L·atm/(mol·K)) (275 K)
n ≈ 0.3709 mol
Now we can calculate the work:
W = - (0.3709 mol) (0.0821 L·atm/(mol·K)) (275 K) ln(7.04 L / 3.64 L)
Using a calculator:
W ≈ - 24.062 J
Therefore, the work for Path A is approximately -24.062 J (remember to include the negative sign since work is done on the system during expansion).
To calculate the work for Path A, you correctly used the equation W = -PΔV, where P is the pressure and ΔV is the change in volume. However, the pressure you used (2.68 atm) is incorrect.
For an isothermal, reversible expansion or compression, we can use the ideal gas law, PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature.
In this case, since the gas sample is at a constant temperature (275 K), the equation becomes P₁V₁ = nRT. We can rearrange it to find n, the number of moles of gas: n = P₁V₁ / RT.
Given:
V₁ = 3.64 L (initial volume)
P₁ = 2.68 atm (initial pressure)
T = 275 K (temperature)
R = 0.0821 L·atm/(mol·K) (ideal gas constant)
First, calculate the number of moles, n:
n = (P₁V₁) / (RT)
= (2.68 atm)(3.64 L) / (0.0821 L·atm/(mol·K))(275 K)
= 0.3598 mol
For an isothermal, reversible expansion, the work is given by the equation:
W = -nRT·ln(V₂/V₁)
V₁ = 3.64 L (initial volume)
V₂ = 7.04 L (final volume)
n = 0.3598 mol (number of moles)
R = 0.0821 L·atm/(mol·K) (ideal gas constant)
T = 275 K (temperature)
Now, substitute these values into the equation to calculate the work, W:
W = -0.3598 mol · (0.0821 L·atm/(mol·K)) · ln(7.04 L/3.64 L)
Calculating the natural logarithm and doing the calculation:
W ≈ -0.3598 mol · (0.0821 L·atm/(mol·K)) · ln(1.93)
W ≈ -0.3598 mol · (0.0821 L·atm/(mol·K)) · 0.65775
W ≈ -0.019 L·atm · (0.243 L·atm/(mol·K))
W ≈ -0.0046 L·atm²/mol·K
To convert L·atm²/mol·K to joules (J):
1 L·atm = 101.3 J
1 K = 1 J
W ≈ -0.0046 L·atm²/mol·K · (101.3 J/L·atm) · (1 J)
W ≈ -0.465 J
The work for Path A is approximately -0.465 J.