ThermodynamicsHard
Question
Two moles of helium gas are taken over the cycle ABCDA, as shown in the P - T diagram.

The net work done on the gas in the cycle ABCDA is

The net work done on the gas in the cycle ABCDA is
Options
A.Zero
B.276 R
C.1076 R
D.1904 R
Solution
Net work done in a cycle = WAB + WBC + WCB + WBA
= 400 R + 2 × 2.303 × 500 R ln 2 - 400R - 414 R
= 1000R x ln 2 - 600R x ln 2 = 400R x ln 2 = 276R
= 400 R + 2 × 2.303 × 500 R ln 2 - 400R - 414 R
= 1000R x ln 2 - 600R x ln 2 = 400R x ln 2 = 276R
Create a free account to view solution
View Solution FreeMore Thermodynamics Questions
When an enclosed perfect gas is subjected to an adiabatic process:...10 mole of an ideal gas is undergoing the process shown in the figure. The heat involved in the process from $P_{1}$ to ...The relation Cp − Cv = R/J implies :-(J−Joule constant)...Even Carnot engine cannot give 100% efficiency because we cannot...Work done depends in gaseous process....