Solving Hydrogen Gas Problem: Ktr & Temp Calculations

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lesodk
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I need to solve this problem:
A gas of 5.00 L hydrogen with temperature 300 K and pressure 1.013 x 10^5 Pa.
The molar mass of Hydrogen is 2.016 g/mol.

a) find the translatinal kinetic Ktr energy of the gass

b) if the gass is contained in a container which is moving 300 m/s what is the new kinetic energy of the gass and what is the temperature?
 
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Hi - welcome to PF.
The note you got when you posted the question said that you had to show us what you had done - otheriwse it's just us telling you the answer.

1, S what is translational KE?
2, Think - if you put a can of hairspray in a train would it become warmer if the train was moving?
 
Last edited:
i meant translational kinetic energy.
 
Sorry - I wasn't making a point about the spelling (I mistyped aswell)
I meant you have to at least lookup the formula and then tell us what you don't understand about it.
 
i use the following for my calculations:

a) K_tr1 =3/2 nRT = 3/2 pV =3/2*1,013*10^5 Pa * 0,005m^3 = 759,75J

b)
nRT=pV <=> n = pV/RT = (1,013*10^5 Pa * 0,005m^3)/(8,314 J/mol*K * 300K) = 0,2 mol

m = n*M = 0,2*2,016g/mol = 4*10^-4 kg

K_tr_2 =1/2mv² = 1/2 *4*10^-4kg*(300m/s)² = 18,4JTotal kinetic energy:
K_tot=K_tr1+K_tr_2 = 759,75J+18,4J=778,15J

K_tot = 3/2nRT <=> T= 2K_tot/3nR <=> T=(2*778,15J)/(3*0,20310 mol*8,314J/mol*K) <=> T= 307K

Does this seem right?