Two adiabatic expansions of n mole of same gas are shown. If `V_B/V_A = V_D / (V_C)xx1/2` then `T_C//T_D` is :

Options

A. `2^(0.4) T_A/T_B`
B. `2^(1.4) T_A/T_B`
C. `T_A/T_B 1/(""_(2)(0.4))`
D. `T_A/T_B`

Answer

A

Explanation

`T_A V_A^(gamma -1) = T_B V_B^(gamma-1) rArr T_A //T_B = ((V_B)/(V_A))^(gamma-1)`
and `T_C V_C^(gamma - 1) = T_D V_D^(gamma -1) rArr T_C //T_D = ((V_D)/(V_C))^(gamma - 1)rArrT_C/T_D = ((2V_B)/V_A)^(gamma-1)= 2gamma-1(T_A)/(T_B) rArr Delta T = (16/5 -1)1/R K = 11/(5R)K``Delta Q = Delta U + W`, Here, Q : U and W all different forms of energy.For isothermal process, `pV` = constant `rArr` slope ` = (dp)/(dV) = (-P)/V`
& for adiabatic process, `P v^(gamma)` = constant `rArr (dp)/(dV) = (- gamma p)/(V) rArr x = gamma`