A sample of gas in a cylinder of volume 4.2 L at 316 K and 2.68 atm expands to 7.21 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 has two steps. In the first step, the gas is cooled at constant volume (4.2 L) until its pressure drops to 1.56 atm. In the sec- ond step, the gas is heated and allowed to expand against a constant external pressure of 1.56 atm until the final volume is 7.21 L. Calculate the work for Path B.
Answer in units of J
To calculate the work done by a gas during an expansion, we can use the formula:
Work = -P * ΔV
Where P is the pressure and ΔV is the change in volume. Let's calculate the work for Path A first.
In Path A, the expansion is isothermal and reversible. This means that the temperature remains constant throughout the process. Since it's an isothermal expansion, we can use the ideal gas law to find the initial and final pressure of the gas.
The ideal gas law is given by:
PV = nRT
Where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant (8.314 J/(mol·K)), and T is the temperature.
Rearranging the equation to solve for n, we get:
n = PV / RT
Since the number of moles remains constant during the expansion, we can calculate the initial number of moles using the given initial conditions. Plugging in the values, we have:
n_initial = (2.68 atm * 4.2 L) / (0.0821 L·atm/(mol·K) * 316 K)
n_initial ≈ 0.368 mol
Now, we can calculate the initial pressure of the gas:
P_initial = (n_initial * R * T_initial) / V_initial
Plugging in the values, we have:
P_initial = (0.368 mol * 8.314 J/(mol·K) * 316 K) / 4.2 L
P_initial ≈ 785.38 J/L
To find the work done during this expansion, we need to calculate the change in volume:
ΔV = V_final - V_initial
= 7.21 L - 4.2 L
= 3.01 L
Finally, we can calculate the work done:
Work_A = -P_initial * ΔV
Plugging in the values, we have:
Work_A = -(785.38 J/L) * 3.01 L
Work_A ≈ -2,363.14 J
Therefore, the work done in Path A is approximately -2,363.14 J.
Now, let's calculate the work for Path B, which consists of two steps.
In the first step, the gas is cooled at constant volume until its pressure drops to 1.56 atm. Since the volume remains constant, the work done in this step is zero.
In the second step, the gas is heated and allowed to expand against a constant external pressure of 1.56 atm. To calculate the work done in this step, we can use the same formula as before:
Work_B = -P * ΔV
Since the pressure is constant at 1.56 atm, we can use this value for P. The change in volume is calculated as:
ΔV = V_final - V_initial
= 7.21 L - 4.2 L
= 3.01 L
Plugging in the values, we have:
Work_B = -(1.56 atm) * 3.01 L
Work_B ≈ -4.7056 atm·L
To convert the units to joules, we need to multiply by the conversion factor: 1 atm·L = 101.325 J.
Work_B = (-4.7056 atm·L) * (101.325 J/atm·L)
Work_B ≈ -475.97938 J
Therefore, the work done in Path B is approximately -475.98 J.