Solving Hydrogen Gas Problem: Ktr & Temp Calculations

AI Thread Summary
The discussion focuses on solving a problem involving the translational kinetic energy of hydrogen gas. The initial kinetic energy is calculated using the formula K_tr = 3/2 nRT, resulting in approximately 759.75 J. When the gas is in a moving container at 300 m/s, the additional kinetic energy is calculated to be about 18.4 J, leading to a total kinetic energy of 778.15 J. The new temperature of the gas, after considering the total kinetic energy, is determined to be approximately 307 K. The calculations and reasoning presented appear to be accurate based on the provided formulas and values.
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?
 
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