A sample consisting of 1.00 mol Ar is expanded isothermally at 0°C
from 22.4 dm3 to 44.8 dm3 (a) reversibly, (b) against a constant external
pressure equal to the final pressure of the gas, and (c) freely (against zero
external pressure). For the three processes calculate q, w, ΔU, and ΔH.
(Ans. (a) ∆U = ∆H = 0; q = 1.57KJ, w = −1.57KJ, (b) ∆U =
∆H = 0; q = 1.13KJ, w = −1.13KJ, (c) ∆U = ∆H = 0; q = 0, w =
0.)
à) w=-1.57kj, ∆u=0(as isothermal process) and Q= ∆u-w=0-(-1.57kj) =1.57kj
b) w=-1.134kj ,∆u =0and q=1.134kj
c) w=0(as pext. =0) so p*∆v=0=w
∆u=0and q=0
Well, well, well! Let's start with the first scenario, shall we?
(a) When Ar is expanded reversibly, the process is isothermal. This means the temperature remains constant at 0°C. Since the process is reversible, the pressure will also be constant during the expansion. The ideal gas equation, PV = nRT, can be used to calculate the final pressure based on the initial pressure.
Now, let's get into the calculations!
1. q (heat transfer): In an isothermal process, the change in internal energy (ΔU) is zero. Hence, q = 0.
2. w (work done): Since the process is reversible, we can use the formula w = -PΔV. Plugging in the values, w = -Pext × ΔV.
3. ΔU (change in internal energy): As mentioned before, for an isothermal process, ΔU = 0.
4. ΔH (change in enthalpy): In case of an ideal gas, ΔH = ΔU + PΔV. Since ΔU is zero, ΔH = PΔV.
(b) In the second case, the Ar gas is expanded against a constant external pressure equal to the final pressure of the gas.
1. q: As the process is isothermal, q = 0.
2. w: Since the pressure is constant, we can use the formula w = -Pext × ΔV.
3. ΔU: Since the process is isothermal, ΔU = 0.
4. ΔH: Similar to the previous case, ΔH = PΔV.
(c) In the final scenario, the Ar gas is expanded freely against zero external pressure.
1. q: As always, in an isothermal process, q = 0.
2. w: Since the external pressure is zero, the work done is zero as well. So w = 0.
3. ΔU: Isothermal process, again ΔU = 0.
4. ΔH: Same deal here, ΔH = PΔV.
And that's how the cookie crumbles!
To calculate q, w, ΔU, and ΔH for the three different processes, we'll need to use the ideal gas equation and the first law of thermodynamics. Here are the steps to calculate the values for each process:
(a) Reversible Expansion:
For an isothermal expansion of an ideal gas, we can use the equation:
ΔU = q + w
Since the expansion is isothermal, the change in internal energy (ΔU) is zero. Therefore, q = -w.
Now, to calculate q, we can use the ideal gas equation:
PV = nRT
P1V1 = P2V2
Given:
P1 = ? (We need to find this)
V1 = 22.4 dm3
P2 = ? (We need to find this)
V2 = 44.8 dm3
n = 1.00 mol
R = 8.314 J/(mol K) (Universal gas constant)
T = 0°C = 273 K
Rearranging the ideal gas equation, we can solve for P1 and P2:
P1 = (nRT) / V1 = (1.00 mol * 8.314 J/(mol K) * 273 K) / 22.4 dm3
P2 = (nRT) / V2 = (1.00 mol * 8.314 J/(mol K) * 273 K) / 44.8 dm3
Now that we have P1 and P2, we can calculate q and w using the equation q = -w:
q = -w = -PΔV
Substituting the values of P2, V2, and V1, we can find q and w.
To calculate ΔU and ΔH, we need additional information about the process (e.g., whether it is isobaric or adiabatic).
(b) Expansion against a constant external pressure:
If the expansion is against a constant external pressure, the gas does work against the pressure, and we can use the equation:
w = -PextΔV
q = ΔU + w
To calculate q and w, replace Pext with the final pressure of the gas.
To calculate ΔU and ΔH, we again need additional information about the process.
(c) Free Expansion:
For a free expansion against zero external pressure, no work is done on or by the gas since there is no opposing force. Therefore, w = 0.
To calculate q, we can again use the ideal gas equation to determine the final pressure P2. Then substitute P2, V2, and V1 in the equation q = -w to find q.
To calculate ΔU and ΔH, we again need additional information about the process.
Note: The calculations for ΔU and ΔH are process-dependent and require additional information about the system to determine the heat absorbed or released and the change in internal energy or enthalpy.