Mass of used hydrogen in an isothermal process

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Homework Help Overview

The discussion revolves around calculating the mass of hydrogen used in an isothermal process, given specific conditions such as temperature, volume, and pressure differential. The subject area includes thermodynamics and gas laws.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply a formula involving molar mass, temperature, volume, and pressure differential to find the mass of hydrogen used. Participants question the appropriateness of the units and the values used in the calculations.

Discussion Status

Participants are actively engaging with the original poster's approach, confirming the use of the equation while also pointing out potential issues with unit conversions and specific values. There is a focus on clarifying the calculations rather than reaching a definitive conclusion.

Contextual Notes

There are discussions about the correct interpretation of the molar mass and the gas constant, as well as the volume conversion from liters to cubic meters. The original poster's initial result of 4 kg is questioned, indicating a need for further verification of the calculations.

Uku
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Homework Statement



I have hydrogen in a 12 L tank, at T=15 C. Some of it is used, the T = const. and Δp=0.4 MPa,
the molar mass of hydrogen is M=2*10^-3 J/(mol*K). Find the mass of the used hydrogen.

Homework Equations



Am I wrong in simply using:

\Delta m=\frac{M}{RT}V\Delta p ?

That gives roughly 4 kg's..

Thanks,
U.
 
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Hi Uku,

The equation you have used is appropriate. You could have gone wrong in the units, though.
 
By "could have" you mean you put in the numbers and got a different answer?
 
No, I didn't put in the numbers. I was just confirming the equation use :smile:

Though, 4 kg does seem a bit too much...:rolleyes:
 
Uku said:
By "could have" you mean you put in the numbers and got a different answer?

The answer is different from yours. What number did you use for the volume? And the molar mass is 2*10-3 kg/mol, not J/(mol*K):-p
 
Well the pressure differential is 0.4 MPa..
 
The pressure is all right, but what number have you plugged in for volume?ehild
 
0.012 m3
 
Well, then plugging in all data,
Δm=(2*10-3*0.012*0.4*10-6)/(8.3145*288)=9.6/2394 how can it be 4 kg? :confused:

ehild
 
  • #10
I messed up on R, since I wrote R=8.314*10^{-3}\frac{kJ}{K} when trying to write R in terms of kJ.
 
  • #11
That I did not guess. Why in kJ-s?

But You got the correct result at the end now? :biggrin:

ehild
 

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