Thermodynamics MCQs for NEET — Physics Questions with Answers

In thermodynamics, the work done by the system on the surroundings is considered positive, while the work done on the system by the surroundings is considered negative. This is a convention used to maintain consistency in energy calculations.

For free expansion Q = W = O T = const $ \mu \therefore Eint = 0 $

PV $ \nu $ = Const (an adiabatic process) PV = mRT $ \therefore V { RT \over P } $

AB --> Constant P, increasing V, increasing T BC --> Constant T, increasing V, decreasing P CD --> Constant V, decreasing P, decreasing T DA --> Constant T, decreasing V, increasing P Also BC is at highes temperature than AD

$ PV ^ {\gamma} $ = const [ adiabatic process ] ideal gas PV = RT (m = 1) $ \therefore V = { RT \over P } $ $ \therefore P \left( { RT \over P } \right) ^ {\gamma } = const $ $ { T^{ \gamma} \over P^{ \gamma -1}} =const $

$ W = RT loge {V_2 \over V_1 } = 2-3 \left( { M \over Mo} \right) RT log _{10} { V_2 \over V_1 } $

PV = constant (isothermal Process) $ P \triangle V - V \triangle P = 0 $ $ { \triangle P \over P } = { - \triangle V \over V} $

$ PV ^ {\gamma} $ = Constant (an adiabatic Process) $ Pr V^{ \gamma -1 } \triangle V + V^ {\gamma} \triangle P = 0 $

$ { \left( \triangle P \over P \right)_{adiabatic } \over \left( \triangle P \over P \right)_{isothermal}}$ $ ={ - \gamma { \triangle V \over V } \over - { \triangle V \over V } }= g $