3 moles of an ideal monoatomic gas at 27 C and 1 atm pressure are subjected to a reversible adiabatic Compression to half the initial volume calculate Q , W and U in calories?

Question

dls · Answer

U=5400J?

anonymous · Answer

Nope...)

anonymous · Answer

$$\Delta U = Q + W$$

Here Q=0 Since Adiabatic Process

So

$$\Delta U = W$$

anonymous · Answer

$$\frac{ T1 }{ T2}=(\frac{ V2 }{ v1 })^{\gamma -1}$$

kainui · Answer

This problem looks like you'll need to use gamma, do you remember the relation?

kainui · Answer

Yes exactly.

kainui · Answer

Now go through the fun of converting everything around to SI units and then calories and other pointless nonsense lol.

anonymous · Answer

dW = PdV
You can find V from PV=nRT, I believe.

kainui · Answer

You can use that to calculate the initial volume.

anonymous · Answer

May be Finding T2 and Solving Will be Easy...i think..

anonymous · Answer

V_2 = 0.5V_1

kainui · Answer

U=CvdT right? I think you have a way of finding Cv for an ideal gas, just look above and then you have a change in temperature to plug in after the "integration" so you really don't need to go very far.

anonymous · Answer

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