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 solve part (C) of the problem, we can use the ideal gas law and the work done equation for a gas undergoing isothermal expansion.
In part (b), we found that the work done by the gas during the expansion was 9.117 J. This work done equation is given by:
W = -PΔV
where W is the work done, P is the pressure, and ΔV is the change in volume.
We can rearrange this equation to solve for ΔV:
ΔV = -(W / P)
In part (C), the gas is allowed to expand unchecked until its pressure is equal to the external pressure. This means that the final pressure of the gas will be equal to the external pressure, which is 0.20 atm in this case.
To find the final volume of the gas before it stops expanding, we can use the ideal gas law equation:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
Since the temperature is constant (given as 20 degrees Celsius), we can set up the following equation:
(P1)(V1) = (P2)(V2)
where P1 and V1 are the initial pressure and volume, and P2 and V2 are the final pressure and volume.
Using the given information, we have:
(0.050 L)(1 atm) = (V2)(0.20 atm)
Solving for V2, we find:
V2 = (0.050 L)(1 atm) / (0.20 atm)
V2 = 0.250 L
Therefore, the final volume of the gas before it stops expanding is 0.250 L.
To find the work done during the expansion, we can use the work done equation again:
W = -PΔV
Substituting the values, we have:
W = -(0.20 atm)(0.250 L - 0.050 L)
W = -(0.20 atm)(0.200 L)
W = -0.040 J
Note that the work done is negative because the gas is expanding against an external pressure.
Therefore, the work done during the expansion until the gas reaches the external pressure is -0.040 J and the final volume before it stops expanding is 0.250 L.
To calculate the work done by a gas in an expansion process, you can use the formula:
Work = -Pext * ΔV
where Pext is the external pressure and ΔV is the change in volume of the gas.
In part (b), you have already calculated the work done to be 9.117 J. Now let's move on to part (c) of the question.
To find the final volume of the gas before it stops expanding, we need to consider that the gas will continue to expand until its pressure is equal to the external pressure (0.20 atm in this case).
To find the final volume, we can make use of the ideal gas law:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.
We already know the initial volume, pressure, and number of moles, so we can rearrange the ideal gas law equation to solve for the final volume:
Vf = (Pi * Vi * Tf) / (ni * Ti)
where Vf is the final volume, Pi is the initial pressure, Vi is the initial volume, Tf is the final temperature, ni is the initial number of moles, and Ti is the initial temperature.
Since the problem states that the temperature remains constant, we can use the initial temperature value for Ti and Tf. Therefore, the equation simplifies to:
Vf = (Pi * Vi) / (ni)
Plugging in the values, we have:
Vf = (0.050 L * 0.20 atm) / (0.020 mole)
Vf = 0.50 L
Therefore, the final volume of the gas before it stops expanding is 0.50 L.
To find the work done during this expansion, we can once again use the formula:
Work = -Pext * ΔV
In this case, since the external pressure is equal to the pressure of the gas, the work done will be zero. This is because if the pressure of the gas is equal to the external pressure, there is no pressure difference and therefore no work is done in expanding or compressing the gas.
So, in part (c) of the question, the final volume before the gas stops expanding is 0.50 L, and the work done by the gas is zero.