Question

One mole of an ideal gas does 3000 J of work on its surroundings as it expands isothermally to a final pressure of 1.00 atm and volume of 250 L. What is the initial volume and the temperature of the gas?

Answer 1

First, let's note a few important conditions:

• Isothermal means the temperature is held constant, so `color(green)(DeltaT = 0)`.

In these conditions, there is no change in enthalpy `DeltaH` nor any change in internal energy `DeltaU`. Not sure if we need it, but we should know that `color(blue)(DeltaH = DeltaU = 0)` for an ideal gas if `DeltaT = 0`.

• We are given final pressure but not initial pressure, and final volume but not initial volume. So, we would probably have to manipulate the Ideal Gas Law `PV = nRT` for `T`, and maybe something to do with the work function for `V_1`.

REVERSIBLE WORK VS VOLUME

Now, we should know that reversible (efficient) work is defined as:

`\mathbf(w_"rev" = -intPdV)`

For this problem, the depiction in a PV-diagram goes like this:

However, since we do not know `DeltaP` and the equation had assumed that `DeltaP = 0` (note that we are integrating with respect to volume, not pressure), we have use the Ideal Gas Law to rewrite the above equation as:

`w_"rev" = -int_(V_1)^(V_2)(nRT)/VdV`

Since `DeltaT = 0`, `nRT` is a constant. Let's pull it out of the integral.

`w_"rev" = -nRTint_(V_1)^(V_2)1/VdV`

`= -nRTln|V_2/V_1|`

At this point, the shape of the PV-diagram curve looks sensible; it resembles the shape of a `-lnx` curve.

THE INITIAL VOLUME FOR THE EXPANSION

We are given the work, so we should be able to solve for the initial volume, `V_1`, using this equation. Since expansion work has been done by the gas, meaning that `V_2 > V_1`, we know that `color(green)(w_"rev" < 0)`, numerically.

That makes sense because `ln|V_2/V_1|` when `V_2/V_1 > 1` is positive. Now let's get an expression for `V_1`.

`-w_"rev"/(nRT) = ln|V_2/V_1|`

`e^(-w_"rev""/"nRT) = V_2/V_1`

`e^(w_"rev""/"nRT) = V_1/V_2`

`color(green)(V_1 = V_2e^(w_"rev""/"nRT))`

However, we do not know the temperature yet, so we'll have to put off calculating the initial volume for a bit longer.

THE TEMPERATURE FOR THE EXPANSION

Something we do know is that the temperature remained constant, so `T_2 = T_1`. Thus, we have these two relationships that cover the initial and final states:

`P_1V_1 = color(blue)(nR)T`
`color(blue)(P_2V_2) = color(blue)(nR)T`

We know the values of what is in blue, which is enough.

`color(blue)(T) = (P_2V_2)/(nR) = (("1 atm")("250 L"))/(("1 mol")("0.082057 L"cdot"atm/mol"cdot"K"))`

`=` `color(blue)("3046.66 K")`

Now we can find the initial volume:

`color(blue)(V_1) = ("250 L")e^((-"3000 J")"/"[("1 mol")("8.314472 J/mol"cdot"K")("3046.66 K")]`

`=` `color(blue)("222.08 L")`

(As an aside, if you were curious, the initial pressure was about `"1.126 atm"`, from the Ideal Gas Law.)