One mole of an ideal gas is allowed to expand isothermally at -29. 3degrees celcious from a volume of 4. 0L and apressure of 5. 0atm to avolume of 10. 0L and apressure of 2. 0atm.
i. Draw a p-v diagram for the process.
ii. Determine the work done by the fasduring expansion.
To solve this problem, we will use the ideal gas law equation and the formula for work done.
i. Draw a p-v diagram for the process:
A p-v (pressure-volume) diagram represents the changes in pressure and volume of the gas during the process. The x-axis represents the volume (V), and the y-axis represents the pressure (P).
In this case, the gas is expanding isothermally (at a constant temperature). Since the pressure and volume are changing, we can draw a rectangular plot on the p-v diagram.
Starting at point A with a volume of 4.0L and a pressure of 5.0atm, draw a horizontal line extending to point B with a volume of 10.0L. Then, draw a vertical line extending down from point B to intersect the x-axis.
Your p-v diagram should have a rectangular shape, with the horizontal line representing the expansion of volume from 4.0L to 10.0L, and the vertical line representing the decrease in pressure from 5.0atm to 2.0atm.
ii. Determine the work done by the gas during expansion:
The work done by the gas during expansion can be found using the formula:
Work (W) = -P * ΔV
Where P is the average pressure during expansion and ΔV is the change in volume.
In this case, the average pressure (P) is the average of the initial and final pressures, so we can calculate it as follows:
P = (5.0atm + 2.0atm) / 2 = 3.5atm
The change in volume (ΔV) is the final volume minus the initial volume:
ΔV = 10.0L - 4.0L = 6.0L
Now we can substitute the values of P and ΔV into the formula to find the work done by the gas:
W = -(3.5atm) * (6.0L)
W = -21.0 atm.L
The work done by the gas during expansion is -21.0 atm.L. Note that the negative sign indicates that work is done on the gas during expansion.
Hope this helps!