JEE Advanced2019PhysicsThermodynamicsActual
A mixture of ideal gas containing 5 moles of monatomic gas and 1 mole of rigid diatomic gas is initially at pressure P 0 , volume V 0 and temperature T 0 . If the gas mixture is adiabatically compressed to a volume V 0 4 , then the correct statement(s) is/are, (Given 2 1.2 = 2.3 ; 2 3.2 = 9.2 ; R is gas constant)
Options
- AThe final pressure of the gas mixture after compression is in between 9 P 0 and 10 P 0 .
- BThe average kinetic energy of the gas mixture after compression is in between 18 R T 0 and 19 R T 0 .
- CThe work W done during the process is 13 R T 0 .
- DAdiabatic constant of the gas mixture is 1.6 .
Correct answer
A. The final pressure of the gas mixture after compression is in between 9 P 0 and 10 P 0 .
Step-by-step solution
Here,   n 1 = 5 , C P 1 = 5 2 R , C V 1 = 3 2 R n 2 = 1 ,   C P 2 = 7 2 R , C V 2 = 5 2 R γ mix = n 1 C P 1 + n 2 C P 2 n 1 C V 1 + n 2 C V 2 = 8 5 = 1.6 Work done, W = P 1 V 1 - P 2 V 2 γ - 1 ,   In   adiabatic   process Now, for adiabatic compression P 1 V 1 r = P 2 V 2 r ⇒ P 0 V 0 8 / 5 = P 2 V 0 4 8 / 5 ⇒ P 2 = 9.2 P 0 ⇒ A is correct. Now, W = P 0 V 0 - 9.2 P 0 V 0 4 3 / 5 = - 13 R T 0 ⇒ C is correct ∴ W = 13   R T 0 Again T 1 V 1 γ - 1