Consider a system with ideal gas only.
External P > internal P. System is compressed.
Initial temp = final temp. Tᵢ = Tf.
What work is done?
PV = nRT.
Pressure is increasing. Volume is decreasing.
1) Work is done on our system, since we are compressing?
2) Work is considered PV work and system energy decreases?
Ideal gas and work?
2018-01-25 3:40 pm
回答 (2)
2018-01-25 5:42 pm
✔ 最佳答案
Initially, External P > Internal PThe system is compressed from Vi to Vf by constant P (external P), until External P = Internal P = P
Work done on the system
= -ΔPV
= -PΔV
= -P(Vf - Vi) > 0
1)
Yes, work is done on the system since the system is compressed by external pressure.
2)
Work is done on the system by the surroundings. This means that energy is flow from the surroundings to the system. Therefore, the energy of system increases.
2018-01-25 7:45 pm
(1) Yes, work is done ON the system, W = p delta-V.
(2) Normally the internal energy would increase, but with Ti = Tf, apparently the internal energy has remained the same, so some heat must have escaped from the system.
(2) Normally the internal energy would increase, but with Ti = Tf, apparently the internal energy has remained the same, so some heat must have escaped from the system.
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