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Question: A certain mass of gas at \[273\,K\] is expanded to \[81\] times its volume under adiabatic condition...

A certain mass of gas at 273K273\,K is expanded to 8181 times its volume under adiabatic conditions. γ=1.25\gamma = 1.25 for the gas then its final temperature is:
A. 182C - {182^ \circ }C
B. 0C{0^ \circ }C
C. 235C - {235^ \circ }C
D. 91C - {91^ \circ }C

Explanation

Solution

The adiabatic expansion of the gas is the expansion in which the change in heat energy is zero. The equation of state for adiabatic processes states that the pressure times the volume power to the γ\gamma is constant.

Formula used:
The equation of state for adiabatic process is given by,
PVγ=KP{V^\gamma } = K
where PP is pressure of the gas VV is the volume of the gas γ\gamma is the ratio of molar specific heat and KK is some constant.
Equation of state for ideal gas is given by,
PVT=K\dfrac{{PV}}{T} = K
where, PP is pressure of the gas , VV is the volume of the gas TT is the absolute temperature of the gas and KK is universal gas constant.

Complete step by step answer:
We know that in adiabatic processes the loss or gain of heat energy by the system is zero. By the second law of thermodynamics, dQ=0dQ = 0 is the loss or gain of heat energy. In adiabatic process the equation of state is given by,
PVγ=KP{V^\gamma } = K

Now, we know that equation of state for ideal gas law states that the ratio of pressure times volume to absolute temperature is a constant of motion i.e.
PVT=K\dfrac{{PV}}{T} = K
So, putting the value of pressure in the adiabatic process state equation we have,
TVγ1=KT{V^{\gamma - 1}} = K
Now, for two states of the system we can write,
T1V1γ1=T2V2γ1{T_1}{V_1}^{\gamma - 1} = {T_2}{V_2}^{\gamma - 1}

Now, here we have given, initial temperature T1=273{T_1} = 273 final volume in terms of initial volume V2=81V1{V_2} = 81{V_1} and value of γ=1.25\gamma = 1.25. We have to find the final temperature,
So, putting the values we have,
273×V11.251=T2×(81V1)1.251273 \times {V_1}^{1.25 - 1} = {T_2} \times {(81{V_1})^{1.25 - 1}}
T2=273(81)0.25\Rightarrow {T_2} = \dfrac{{273}}{{{{(81)}^{0.25}}}}
T2=2733\Rightarrow {T_2} = \dfrac{{273}}{3}
T2=91\therefore {T_2} = 91
Hence the final temperature of the gas is 91K91K or (91273)C=182C{(91 - 273)^ \circ }C = - {182^ \circ }C.

Hence, the correct answer is option A.

Note: In adiabatic process since the exchange of energy is zero so the work done by the system is achieved by losing some of the internal energy of the gas. The ideal gas law is valid only for ideal gases while the adiabatic equation of state is valid also for real gases.