Mass of used hydrogen in an isothermal process

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SUMMARY

The discussion focuses on calculating the mass of used hydrogen in an isothermal process involving a 12 L tank at a constant temperature of 15°C and a pressure differential of 0.4 MPa. The appropriate equation used is Δm = (M/RT)VΔp, where M is the molar mass of hydrogen (2 x 10-3 kg/mol), R is the ideal gas constant (8.3145 J/(mol·K)), and T is the temperature in Kelvin. The correct calculation yields approximately 9.6 grams of used hydrogen, correcting earlier miscalculations related to unit conversions and the value of R.

PREREQUISITES
  • Understanding of the ideal gas law and its applications
  • Familiarity with unit conversions, particularly between J and kJ
  • Knowledge of thermodynamic principles, specifically isothermal processes
  • Basic proficiency in algebra for manipulating equations
NEXT STEPS
  • Study the ideal gas law and its implications in thermodynamic processes
  • Learn about unit conversions in thermodynamics, especially between different energy units
  • Explore isothermal process calculations in greater detail
  • Investigate the properties of hydrogen as a gas, including its molar mass and behavior under varying pressures
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Students studying thermodynamics, engineers working with gas systems, and anyone involved in chemical process calculations will benefit from this discussion.

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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