Question

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?

Answer 1

@Vincent-Lyon.Fr

Answer 2

for adiabatic process
du=-dw

Answer 3

so find the work done for this adiabatic process.

Answer 4

w= ((p1v1 - p2v2)/(gamma -1)

Answer 5

gamma= cp/cv

Answer 6

for monoatomic gamma= 5/3

Answer 7

does it help u?

Answer 8

Q=0 in an adiabatic process.

Answer 9

as u know that in an adiabatic process there is no heat exchange.

Answer 10

Adiabatic => dQ = 0 => dU = dW. You can get new V1 from pV = nRT, then use the same equation and V2 = 1/2 V1 to get T2, then use the energy equation of state for a monatomic ideal gas, U = 3/2 nRT to calculate dU. Alternatively you can integrate pdV to find dW, using the canonical relationship between p and V during an adiabatic process, pV^gamma = constant.

Answer 11

I agree with what is written above.
Writing dU = δQ + δW or dU = δQ - δW is only a matter of convention.
In France, we use the former one.
You can prove that W = \(\Large \frac {1}{\gamma - 1}\large (P_2 V_2-P_1V_1)\) by integrating elementary work, but the most elegant way is that suggested by Carl_Pham.

In your case, work out final pressure using \(PV^\gamma=cst\), then use formula above to get W.