Formula for the Internal Energy of an Ideal Gas???

[Jumb0]

Ok, so I thought the equation was U = (3/2)nRT , but I have come across a Kaplan discrete that would have me believe that it is actually U = (3/2) n deltaT ...This does not seem like a proper rearrangement of the formula to me. They have completely omitted the gas constant.

The question was:

"If 2 moles of an ideal gas absorb 900 J of heat energy while they perform 600 J of work on the surroundings, what is the change in temperature of the gas?"

So you start with W= Q- W = 900 J - 600 J = 300 J

I'm okay up to this point...but then the solution says that the change in temperature is found by:

delta U = (3/2) n deltaT

Therefore, delta T = 2 delta U / 3 n = 2(300) / 3(2) = 100 J

BUT WHAT ABOUT THE GAS CONSTANT???

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[DoctorInASaree]

The internal energy of an ideal gas is only dependent on temperature. I wouldn't worry about that question though; I learned this in an upper level thermodynamics course It would be highly unlikely for such a question to appear on the MCAT.

 

[kraskadva]

Ok, so I thought the equation was U = (3/2)nRT , but I have come across a Kaplan discrete that would have me believe that it is actually U = (3/2) n deltaT ...This does not seem like a proper rearrangement of the formula to me. They have completely omitted the gas constant.

The question was:

"If 2 moles of an ideal gas absorb 900 J of heat energy while they perform 600 J of work on the surroundings, what is the change in temperature of the gas?"

So you start with W= Q- W = 900 J - 600 J = 300 J

I'm okay up to this point...but then the solution says that the change in temperature is found by:

delta U = (3/2) n deltaT

Therefore, delta T = 2 delta U / 3 n = 2(300) / 3(2) = 100 J

BUT WHAT ABOUT THE GAS CONSTANT???

That's because it's not a rearrangement, it's a derivation. And this is something I'm doing in PChem, never in Gen chem, so I doubt it would show up on the MCAT. Also, it doesn't show up on the AAMC Topic Lists...