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For free expansion of the gas which of the following is true ?
For free expansion Q = W = O T = const $ \mu \therefore Eint = 0 $
For an adiabatic process involving an ideal gas
PV $ \nu $ = Const (an adiabatic process) PV = mRT $ \therefore V { RT \over P } $
$ \mu $ moles of gas expands from volume $ V_1 to V_2 $ at constant temperature T. The work done by the gas is
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
One mole of an ideal gas $ {C_p \over C_v} = \gamma $ at absolute temperature $T_1$ is adiabatically compressed from an initial pressure $P_1$ to a final pressure $P_2$ . The resulting temperature $T_2$ of the gas is given by
$ 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 $
In anisothermal reversible expansion, if the volume of 96J of oxygen at $ 27 ^\circ $ is increased from 70 liter to 140 liter, then the work done by the gas will be
$ W = RT loge {V_2 \over V_1 } = 2-3 \left( { M \over Mo} \right) RT log _{10} { V_2 \over V_1 } $
For an iso thermal expansion of a Perfect gas, the value of $ { \triangle P \over P } $ is equal t o
PV = constant (isothermal Process) $ P \triangle V - V \triangle P = 0 $ $ { \triangle P \over P } = { - \triangle V \over V} $
For an adiabatic expansion of a perfect gas, the value of $ { \triangle P \over P} $ is equal to
$ PV ^ {\gamma} $ = Constant (an adiabatic Process) $ Pr V^{ \gamma -1 } \triangle V + V^ {\gamma} \triangle P = 0 $
If r denotes the ratio of adiabatic of two specific heats of a gas. Then what is the ratio of slope of an adiabatic and isothermal P -->V curves at their point of intersection ?
$ { \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 $
Work done permol in an isothermal? change is
$ W = mRT ln { V_2 \over V_1 } = R T log _e { V_2 \over V_1 } $
The isothermal Bulk modulus of an ideal gas at pressure P is
isothermal bulk modulus B = P
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