Pressure/temperature energy question

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  • #26
OmCheeto
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in the equation

n= pV/RT

isn't pV divided by the quantity R times T?

pV = nRT

therefore

n = (pV)/(RT) <--- correct equation

n <> (1.01325x10^6)(2.6x10^-1)(273) / (8.31) = 8654691.336
 
  • #27
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ok. I'm still trying to understand the process from this point. So...

now I multiply by 2 because Hydrogen gas has 2 moles and then convert into kg. After that I plug the resulting value of n into

Q = n Cv delta T

Is this correct?
 
  • #28
kuruman
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pV = nRT

therefore

n = (pV)/(RT) <--- correct equation

n <> (1.01325x10^6)(2.6x10^-1)(273) / (8.31) = 8654691.336
This is not correct. T = 273 belongs in the denominator.
 
  • #29
kuruman
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ok. I'm still trying to understand the process from this point. So...

now I multiply by 2 because Hydrogen gas has 2 moles and then convert into kg. After that I plug the resulting value of n into

Q = n Cv delta T

Is this correct?
Yes, but you need to calculate n correctly. See my posting above.
 
  • #30
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I've found another formula in the book that says:

Q = 3/2 n R delta T Is this the same formula as Q = n Cv Delta T just rewritten?


I've found the number of moles = 8654691.336

multiplying by 2 =17309382.67 I get the mass in grams

converting to kg =1.730938267 x 10^7

Plugged into the formula Q=nCvT

Q=(1.7310938267)(10.144)(28)=4916418579 This seems way too big. What am I missing?


Using the new formula I found in the book I get:

(3/2)(1.731 x 10^7)(10.144)(28)= 78911.33242

when converted to kg = 78.9 (which is the answer many of my classmates got correct, their original numbers were slightly different)

thanks for your help.
 
  • #31
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I've recalculated for n and the results are as follows:

n=2.3225 x 10^-1

plug in to the equation

Q = (2.3225 x 10^-1)(10.144)(28)
= 65.966432
= 65.97 when rounded

This is the answer! Thank you very much for all your help---You Rock!
 
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