Question

Question #8cf3a

Answer 1

Via `PV=nRT`:
`T_1~~334K`
`Pa_2~~3.28*10^6Pa=3.28MPa`

Via `(P+(n^2a)/V^2)(V-nb)=nRT`:
`T_1~~286K`
`P_2~~2.76*10^6Pa`

Explanation:

Assuming this behaves as an idea gas, we have the equation:
`PV=nRT`, where:

• `P` = Pressure (`Pa`)
• `V` = Volume (`m^3`)
• `n` = Number of moles (`mol`)
• `R` = Gas constant (`~8.31Jcolor(white)(l)mol^(-1)color(white)(l)K^(-1)`)
• `T` = Temperature (`K`)

To find `T` we can rearrange to get `T=(PV)/(nR)`

`P=33MPa=33*10^6Pa`
`V=3000mm^3=3*10^(-6)m^3`
`n=m/M_r=1/(2*14)=1/28mol`
`R=8.31Jcolor(white)(l)mol^(-1)color(white)(l)K^(-1)`

`T_1=((33*10^6)(3*10^(-6)))/(1/28*8.31)=333.5740072K=334K`

`T_1~~334K`

To find the new pressure at `30144mm^3`, we can use: `P=(nRT)/V`

`P=(1/28*8.31*((33*10^6)(3*10^(-6)))/(1/28*8.31))/(3.0144*10^(-5))`

`=(cancel(1/28*8.31)*((33*10^6)(3*10^(-6)))/(cancel(1/28)*cancel(8.31)))/(3.0144*10^(-5))`

`=((33*10^6)(3*10^(-6)))/(3.0144*10^(-5))~~3.28*10^6Pa=3.28MPa`

`P_2~~3.28*10^6Pa=3.28MPa`

Now to compare this with Van der Waals equation of state:
`(P+(n^2a)/V^2)(V-nb)=nRT`:

The only new values are `a` and `b` where:

• `a` = the amount of attraction between particles (`Jcolor(white)(l)m^2color(white)(lmol^(-2)`)
• `b` = volume of a gas per mole (`m^3color(white)(l)mol^(-1)`)

`a` and `b` are constants for specific gases.

For `N_2`, `a=0.137Jcolor(white)(l)m^2color(white)(l)mol^(-2)` and `b=3.87*10^(-5)m^3color(white)(l)mol^(-1)`

To solve for `T` we use:
`T=((P+(n^2a)/V^2)(V-nb))/(nR)`

`=(((33*10^6)+0.137(1/(28(3*10^(-6)))^2))(3*10^(-6)-(3.87*10^(-5))/28))/((8.31/28))`

Which after calculating comes out as `T_1=285.73397349318~~286K`

Now to solve for the new pressure at a constant `T, n, R, a` and `b`

`(P_2+0.137/(28(3.0144*10^(-5)))^2)((3.0144*10^(-5))-(3.87*10^(-5))/28)=(285.73397349318*8.31*1/28)`

`P_2=(285.73397349318*8.31*1/28)/((3.0144*10^(-5))-(3.87*10^(-5))/28)-0.137/(28(3.0144*10^(-5)))^2`

`=312612767184732201015625/113425766518212918`

`=2756100.10654Pa~~2.76*10^6Pa`

Answer 2

I got `"27.563 bar"`, or `2.7563 xx 10^6 "Pa"`.

---

The form of the van der Waals (vdW) equation I know is:

`P = (RT)/(barV - b) - a/(barV^2)`

where:

• `a` and `b` are the vdW constants.
• `P` is the pressure in `"bar"`.
• `barV` is the molar volume in `"L/mol"`.
• `T` is the temperature in `"K"`.
• `R = "0.083145 L"cdot"bar/mol"cdot"K"` is the universal gas constant.

You seem to have given:

• `P_1 = "33 MPa"`
• `V_1 = "3000 mm"^3`
• `T_1 = ? -= T`
  `" "`
• `P_2 = ?`
• `V_2 = "30144 mm"^3`
• `T_2 = T_1 -= T`

Van der Waals constants can be found here or in your text. I see:

`a = "1.370 bar"cdot"L"^2"/mol"^2`
`b = "0.0387 L/mol"`

Therefore, we can solve for the temperature that this gas is at:

`P + a/(barV^2) = (RT)/(barV - b)`

`T = [P + a/(barV^2)][(barV - b)/R]`

Now, we'll need to convert our variables to the proper units.

Initial Variables

`P_1 = 33 cancel"MPa" xx (10^6 cancel"Pa")/(cancel"1 MPa") xx ("1 bar")/(10^5 cancel"Pa") = ul"330 bar"`

`V_1 = 3000 cancel("mm"^3) xx (cancel("1 m")/(1000 cancel"mm"))^3 xx (("10 dm")/(cancel"1 m"))^3`

`= "0.003 dm"^3 = "0.003 L"`

`cancel("1 g N"_2) xx "1 mol"/(28.014 cancel("g N"_2)) = "0.0357 mol N"_2`

`=> barV_1 = "0.003 L"/"0.0357 mols" = ul"0.08404 L/mol"`

Final Variables

`V_2 = 30144 cancel("mm"^3) xx (cancel("1 m")/(1000 cancel"mm"))^3 xx (("10 dm")/(cancel"1 m"))^3`

`= "0.030144 dm"^3 = "0.030144 L"`

`=> barV_2 = "0.030144 L"/"0.0357 mols" = ul"0.84445 L/mol"`

As a result,

`T = ["330 bar" + ("1.370 bar"cdot"L"^2"/mol"^2)/(("0.08404 L/mol")^2)][("0.08404 L/mol" - "0.0387 L/mol")/("0.083145 L"cdot"bar/mol"cdot"K")]`

`=` `ul"285.73 K"`

So, at constant temperature and the new molar volume of `"0.84445 L/mol"`, the new pressure needed is:

`color(blue)(P_2) = (RT)/(barV - b) - a/(barV^2)`

`= ("0.083145 L"cdot"bar/mol"cdot"K" cdot "285.73 K")/("0.84445 L/mol" - "0.0387 L/mol") - ("1.370 bar"cdot"L"^2"/mol"^2)/(("0.84445 L/mol")^2)`

`=` `color(blue)ul("27.563 bar")`