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"#?

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"#.

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")#