1 mole of \[{\text{N}}{{\text{H}}_{\text{3}}}\] gas at \[{27^0}{\text{C}}\] is expanded in reversible adiabatic conditions to make volume 8 times$\left( {{\text{y}} = 1.33} \right)$. Final temperature and work done respectively are:
A.150 K, 909 cal
B.150 K, 400 cal
C.250 K, 1000 cal
D.200 K, 800 cal

Hint:To solve this problem the idea of reversible adiabatic expansion is required. An adiabatic process is one in which there is no exchange of heat from the system to its surroundings neither during expansion nor during compression. In this process if the conditions are made reversible then the work done is considered adiabatic reversible work.
Formula used:
The reversible adiabatic expansion or compression of an ideal gas is given by:
${\text{T}}{{\text{V}}^{{{\gamma - 1}}}}{\text{ = constant}}$
$ \Rightarrow \left( {\dfrac{{{{\text{T}}_{\text{2}}}}}{{{{\text{T}}_{\text{1}}}}}} \right){\text{ = }}{\left( {\dfrac{{{{\text{V}}_{\text{1}}}}}{{{{\text{V}}_{\text{2}}}}}} \right)^{{{\gamma - 1}}}}$
${{\Delta U}}$=${\text{ - n}}{{\text{C}}_{\text{v}}}{\text{dT}}$
Where U is the internal energy, n is the moles, dT is the change in temperature and ${{\text{C}}_{\text{V}}}$ is specific heat capacity at constant volume.

Complete step by step answer:
Given, ${{\text{T}}_1}$=${27^0}{\text{C}} = {\left( {273 + 27} \right)^0}{\text{C = 300 K}}$ \[{{\text{T}}_{\text{2}}}{\text{ = ???}}\]
\[{{\text{V}}_{\text{1}}}\]= V and \[{{\text{V}}_2}\] = 8V.
Therefore, putting these value in the above equation we get,
\[\left( {\dfrac{{\text{T}}}{{300}}} \right) = {\left( {\dfrac{{\text{V}}}{{8{\text{V}}}}} \right)^{\left( {\dfrac{4}{3} - 1} \right)}}\]
$ \Rightarrow {\text{T}} = \dfrac{{300}}{2} = 150{\text{K}}$
So the final temperature of the gas is 150 K.
Now coming to the second part of the problem, from the first law of thermodynamics, we have
${{\Delta U = q + w}}$,
where ${\text{\Delta U}}$ is the change in internal energy, q is the heat supplied to the system, and w is the amount of work done.
For adiabatic processes, the heat change is zero.
So,${{\Delta U = w}}$.
Therefore,
Adiabatic work done= ${{\Delta U}}$=${\text{ - n}}{{\text{C}}_{\text{v}}}{\text{dT}}$ and ${{\gamma = }}\dfrac{{{{\text{C}}_{\text{p}}}}}{{{{\text{C}}_{\text{v}}}}}$
Therefore, \[{\text{w}} = - \left[ {1 \times \dfrac{{\text{R}}}{{\gamma - 1}} \times \left( {150 - 300} \right)} \right]\]
$ \Rightarrow {\text{w}} = - \left[ { - 150 \times \dfrac{2}{{1.33 - 1}}} \right]$= 909 cal
Hence the adiabatic work done is 909 cal.

So the correct answer is option A.

Note:
1. The work done in a reversible adiabatic process is higher than that for irreversible adiabatic processes.
2. In an adiabatic expansion, \[\left( {{{\text{V}}_{\text{2}}}{\text{ > }}{{\text{V}}_{\text{1}}}} \right)\] the gas cools $\left( {{{\text{T}}_{\text{2}}}{\text{ > }}{{\text{T}}_{\text{1}}}} \right)$and in case of compression\[\left( {{{\text{V}}_{\text{2}}}{\text{ < }}{{\text{V}}_{\text{1}}}} \right)\]the gas heats up,$\left( {{{\text{T}}_{\text{2}}}{\text{ < }}{{\text{T}}_{\text{1}}}} \right)$