1.95 mol of an ideal gas at 300 K and 3.00
atm expands from 16 L to 28 L and a final
pressure of 1.20 atm in two steps:
(1) the gas is cooled at constant volume
until its pressure has fallen to 1.20 atm, and
(2) it is heated and allowed to expand
against a constant pressure of 1.20 atm un-
til its volume reaches 28 L.
Which of the following is CORRECT?
1. w = 0 for (1) and w = −1.46 kJ for (2)
2. w = −4.57 kJ for (1) and w = −1.46 kJ
for (2)
3. w = 0 for the overall process
4. w = −4.57 kJ for the overall process
5. w = −6.03 kJ for the overall process
To determine the correct option, we need to calculate the work done in each step and then determine the total work done for the overall process.
Step 1: Cooling at constant volume until the pressure falls to 1.20 atm.
Since the volume remains constant, the work done in this step is given by the equation:
w = -P * ΔV
where P is the pressure and ΔV is the change in volume.
In this case, P = 3.00 atm (initial pressure) and ΔV = 0 L (no change in volume).
Therefore, the work done in step 1 is w = -3.00 atm * 0 L = 0.
Step 2: Heating and allowing expansion against a constant pressure of 1.20 atm until the volume reaches 28 L.
The work done in this step can be calculated using the equation:
w = -P * ΔV
where P is the constant pressure and ΔV is the change in volume.
In this case, P = 1.20 atm (constant pressure) and ΔV = 28 L - 16 L = 12 L.
Therefore, the work done in step 2 is w = -1.20 atm * 12 L = -14.4 L.atm.
Now, let's calculate the total work done for the overall process by adding the work done in each step:
Total work done = work done in step 1 + work done in step 2
= 0 + (-14.4 L.atm)
= -14.4 L.atm
The correct option is 4. w = -4.57 kJ for the overall process.
Note:
To convert the units from L.atm to kJ, we use the conversion factor: 1 L.atm = 101.325 J = 0.101325 kJ.
Therefore, -14.4 L.atm = -14.4 * 0.101325 kJ = -1.45956 kJ, which is approximately -1.46 kJ.