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Grade: 12th pass
Figure shows a cylindrical tube of volume V with adiabatic walls containing an ideal gas. The internal energy of this ideal gas is given by U=3/2nRT. The tube is divided into two equal parts by a fixed diathermic wall (only transfer of energy is possible). Initially, the pressure and the temperature are p1, T1 on the left and p2, T2 on the right. The system is left for sufficient time so that the temperature becomes equal on the two sides.
a) How much work has been done by the gas on the left part?
b) Find the final pressures on the two sides.
c) Find the final equilibrium temperature.
d) How much heat has flown from the gas on the right to the gas on the left?
3 years ago

Answers : (1)

13 Points
							Let n1 , U1 and n2 ,U2 be the no. of moles , internal energy of ideal gas in the left chamber and right chamber respectively.(a) As the diathermic wall is fixed, so final volume of the chambers will be same. Thus, ΔV = 0, hence work done ΔW= PΔV = 0b) Let final pressure in the first and second compartment P1’ and P2’.W
one year ago
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  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Previous Year Exam Questions

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