Question

Hydrogen gas (`"H"_2`) is stored in a tank of volume `"2000 L"` at a temperature of `"653 K"` and `"3 barg"`. Calculate the thermal energy transferred from cooling the hydrogen gas in the tank to `220^@ "C"`?

Neglect changes in kinetic and potential energies and assume ideal gas behaviour, Note: `"1 bar = 100 kPa"`
`R = "0.08206 atm"cdot"K/mol"cdot"K"`
`barC_V ("kJ/mol" cdot °"C") = 2.053 xx 10^(-2) + 7.650 xx 10^(-5) T + 3.288 xx 10^(-9) T^2 - 8.698 xx 10^(-13) T^3`

Answer 1

I get about `"1540 kJ"`, assuming that `"3 barg"` means `"4 bar"` of the usual kind of absolute pressure for an atmospheric pressure of `"1 bar"`.

If I am supposed to use `"3 bar"` as is, it would turn out to be about `"1155 kJ"`.

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First off, `"3 barg"` means `"3 bar"` of gauge pressure, which is the absolute pressure minus the atmospheric pressure. I assume that atmospheric pressure is `"1 bar"`, so that the absolute pressure is...

`P_"abs" = "3 barg" + "1 bar" = "4 bar"`

The mols of gas, assuming ideality, is:

`n = (PV)/(RT)`

`= (4 cancel"bar" xx (100 cancel"kPa")/cancel("1 bar") xx (cancel"1 atm")/(101.325cancel"kPa")cdot 2000 cancel"L")/(0.08206 cancel"L"cdotcancel"atm""/"cancel"mol"cdotcancel"K" cdot 653 cancel"K")`

`= "147.34 mols H"_2`

The tank is cooled from `"653 K"` to `220^@ "C"`, or `"493 K"`. At constant volume (for the tank is rigid!), we know that

`barC_V = ((delbarU)/(delT))_V`

where `barU` is the molar internal energy and `barC_V` is the molar heat capacity at constant volume.

Although the pressure is not constant, since there is no non-PV work being done here, `q_V` is a state function according to the first law of thermodynamics:

`DeltabarU = q_V + w_"PV"`

`= q_V - cancel(int_(barV_1)^(barV_2)PdbarV)^(0)`

So, we can use the change in temperature for this process, knowing that the mols of gas are constant.

The thermal energy transferred is then the change in molar internal energy:

`q_V = DeltabarU = int_(T_1)^(T_2) ((delbarU)/(delT))_VdT`

`= int_(T_1)^(T_2) barC_V(T)dT`

`= int_(T_1)^(T_2) {2.053 xx 10^(-2) + 7.650 xx 10^(-5) T + 3.288 xx 10^(-9) T^2 - 8.698 xx 10^(-13) T^3}dT`

in units of `"kJ/mol"`. Integrating this, we obtain that the change in internal energy is:

`DeltabarU = q_V`

`= {:[2.053 xx 10^(-2)T + 7.650 xx 10^(-5) T^2/2 + 3.288 xx 10^(-9) T^3/3 - 8.698 xx 10^(-13) T^4/4]|:}_("653 K")^("493 K")`

`= [2.053 xx 10^(-2)(493) + 7.650 xx 10^(-5) (493)^2/2 + 3.288 xx 10^(-9) (493)^3/3 - 8.698 xx 10^(-13) (493)^4/4] - [2.053 xx 10^(-2)(653) + 7.650 xx 10^(-5) (653)^2/2 + 3.288 xx 10^(-9) (653)^3/3 - 8.698 xx 10^(-13) (653)^4/4]` `"kJ/mol"`

`= 2.053 xx 10^(-2)(493 - 653) + 7.650 xx 10^(-5) (493^2/2 - 653^2/2) + 3.288 xx 10^(-9) (493^3/3 - 653^3/3) - 8.698 xx 10^(-13) (493^4/4 - 653^4/4)` `"kJ/mol"`

`= -3.2848 - 7.01352 - 0.173850 + 0.0266924` `"kJ/mol"`

`= -"10.45 kJ/mol"`

As a result, the heat flow involved is a magnitude:

`color(blue)(q_V) = "10.45 kJ"/cancel"mol" xx 147.34 cancel"mols"`

`=` `color(blue)("1540 kJ")`