A quantity of 0.020 mole of a gas initially at 0.050 L and 20 degrees Celsius undergoes a constant-temperature expansion until its volume is 0.59 L. Calculate the work done (in joules) by the the gas if it expands (b) pressure of 0.20 atm.

(C) If gas in (b) is allowed to expand unchecked until its pressure is equal to the external pressure, what would its final volume be before it stopped expanding, and what would be the work done?

For part b, I got the work done to be 9.117 J, but I have no idea how to do part C.

To calculate the work done by the gas in part (b), you can use the formula:

Work = -PΔV

Where:
P is the pressure (in atm)
ΔV is the change in volume (in L)

1) Convert the pressure from atm to joules:
1 atm = 101.325 J

Work = -(0.20 atm)(0.59 L - 0.050 L)
Work = -(0.20 atm)(0.54 L)
Work = -0.108 atm·L

2) Convert the work to joules:
1 L·atm = 101.325 J

Work = -0.108 atm·L × (101.325 J / 1 L·atm)
Work = -10.94214 J
Since the work is negative, it means that work is done on the gas.

So, the work done by the gas in part (b) is approximately -10.94214 J or -10.9 J.

Now, let's move on to part (c):

In part (c), it is stated that the gas is allowed to expand unchecked until its pressure is equal to the external pressure. This condition is known as free expansion. In free expansion, no work is done because there is no external pressure acting on the gas.

The final volume can be calculated using the ideal gas law:

PV = nRT

Where:
P is the pressure (in atm)
V is the volume (in L)
n is the number of moles
R is the ideal gas constant (0.0821 L·atm/K·mol)
T is the temperature (in Kelvin)

First, let's convert the initial temperature from Celsius to Kelvin:
20°C = 20 + 273.15 = 293.15 K

Now, we can rearrange the ideal gas law equation to solve for the final volume (Vf):

Vf = (nRT) / Pf

Where Pf is the final pressure in atmospheres.

Substituting the given values:
Vf = (0.020 mol)(0.0821 L·atm/K·mol)(293.15 K) / 0.20 atm
Vf = 24.09145 L

Therefore, the final volume before the gas stops expanding is approximately 24.1 L. As stated earlier, since this is a free expansion, no work is done by the gas in part (c).

To calculate the work done by the gas in both parts of the question, we can use the formula:

Work (W) = -P∆V

Where:
- P is the pressure
- ∆V is the change in volume (final volume - initial volume)

For part (b):

Given:
- Initial volume (V1) = 0.050 L
- Final volume (V2) = 0.59 L
- Pressure (P) = 0.20 atm

Using the formula W = -P∆V, we can calculate the work done:

∆V = V2 - V1 = 0.59 L - 0.050 L = 0.54 L

W = -P∆V = -(0.20 atm)(0.54 L) = -0.108 atm*L

Now, to convert atm*L to joules, we can use the conversion factor 1 atm*L = 101.325 J:

W = -0.108 atm*L * (101.325 J/1 atm*L) = -10.96702 J

Since work is a scalar quantity, we can disregard the negative sign and obtain the final answer as:
W = 10.96702 J (rounded to three significant figures)

For part (c):

In this case, the gas is allowed to expand unchecked until its pressure is equal to the external pressure. So, the final pressure (Pfinal) would be equal to the external pressure value of 0.20 atm.

To determine the final volume, we can use the ideal gas law equation:

P1V1 / T1 = P2V2 / T2

Where:
- P1 and P2 are the initial and final pressures, respectively
- V1 and V2 are the initial and final volumes, respectively
- T1 and T2 are the initial and final temperatures in Kelvin, respectively

Given the conditions:
- Initial volume (V1) = 0.050 L
- Initial pressure (P1) = 0.20 atm
- Final pressure (P2) = 0.20 atm
- Temperature (T) remains constant

Since the temperature is constant, T1 = T2. We can cancel out the temperature terms from the equation, which leaves us with:

P1V1 = P2V2

Now, we can rearrange the equation to solve for V2:

V2 = (P1V1) / P2 = (0.20 atm)(0.050 L) / 0.20 atm = 0.050 L

Therefore, the final volume before the gas stops expanding is 0.050 L.

To calculate the work done, we can use the formula W = -P∆V, as discussed earlier:

∆V = V2 - V1 = 0.050 L - 0.050 L = 0 L

W = -P∆V = -(0.20 atm)(0 L) = 0 J

Hence, the work done by the gas in part (c) is 0 J.

i have no idea..

find by ur self..