Work done in reversible adiabatic process by an ideal gas is given by:
[A] $2.303RT\log \dfrac{{{V}_{1}}}{{{V}_{2}}}$
[B] $\dfrac{nR}{\gamma -1}\left( {{T}_{1}}-{{T}_{2}} \right)$
[C] $2.303RT\log \dfrac{{{V}_{2}}}{{{V}_{1}}}$
[D] None of these.

Hint: Use the first law of thermodynamics here. Heat released or absorbed is zero for an adiabatic process. Use heat capacities in terms of constant volume and pressure to get the correct answer in terms of gamma.

Complete step by step answer:
We know that the first law of thermodynamics states that-
     $\Delta U=q+w$

We should know that $\Delta U$ is the change in internal energy of the system and it is a function of temperature and volume. ‘q’ is the heat absorbed or released and ‘w’ is the work done.
We know that for an adiabatic process, no heat is released from or enters the system. So for an adiabatic process, ‘q’ is zero. Therefore, we can write the above discussion that-
     $\Delta U=w$
Or, we can write that- $\Delta U=pdv$
But we can write that, ${{C}_{v}}={{\left( \dfrac{\partial U}{\partial T} \right)}_{V}}$
The heat capacity of an ideal gas in independent of volume and just depends on temperature, Therefore we can write that- ${{C}_{v}}=\left( \dfrac{dU}{dT} \right)\Rightarrow dU={{C}_{v}}dT$ for an ideal gas.
So, for a reversible adiabatic process of an ideal gas we can write that- $-PdV={{C}_{v}}dT$

But we know that for a mole of an ideal gas,
     $\begin{align}
  & PV=RT=\left( {{C}_{p}}-{{C}_{v}} \right)T \\
 & or,P=\frac{\left( {{C}_{p}}-{{C}_{v}} \right)T}{V} \\
\end{align}$
Therefore, $-\left( {{C}_{p}}-{{C}_{v}} \right)T\dfrac{dV}{V}={{C}_{v}}dT$

We know that, $\dfrac{{{C}_{P}}}{{{C}_{V}}}=\gamma $ therefore, substituting this in the above equation and separating the variables we will get-
     \[\dfrac{dT}{T}+\left( \gamma -1 \right)\frac{dV}{V}=0\]
Now, integrating this we will get- $T{{V}^{\gamma -1}}=constant$
Now, eliminating T we can write that- $P{{V}^{\gamma }}=constant$
So, from these two relations eliminating volume we can write that- ${{P}^{\left( 1-\gamma \right)}}{{T}^{\gamma }}=constant$

Now, let us try to calculate the work done - $W=\int\limits_{{{V}_{1}}}^{{{V}_{2}}}{PdV}$
Now, we have calculated above that $P{{V}^{\gamma }}=constant$=K.
Therefore, we can write that -
     $W=K\int\limits_{{{V}_{1}}}^{{{V}_{2}}}{{{V}^{-\gamma }}dV}=\frac{K}{\gamma -1}\left( {{V}_{1}}^{\left( 1-\gamma \right)}-{{V}_{2}}^{\left( 1-\gamma \right)} \right)=\dfrac{{{P}_{1}}{{V}_{1}}-{{P}_{2}}{{V}_{2}}}{\gamma -1}=\dfrac{R\left( {{T}_{1}}-{{T}_{2}} \right)}{\gamma -1}$

We can understand from the above derivations that work done in a reversible adiabatic process for an ideal gas is $\dfrac{R\left( {{T}_{1}}-{{T}_{2}} \right)}{\gamma -1}$. If we include the number of moles to this, we can write that $\dfrac{nR\left( {{T}_{1}}-{{T}_{2}} \right)}{\gamma -1}$ , where ${{T}_{1}}\text{ and }{{\text{T}}_{2}}$ are the initial and final temperature respectively.
So, the correct answer is “Option B”.

Note: For an ideal gas, the calculated value of $\gamma $ is $\dfrac{5}{3}$ . As we discussed above that $\dfrac{{{C}_{P}}}{{{C}_{V}}}=\gamma $ and for an ideal gas, ${{C}_{P}}=\dfrac{5}{2}$ and ${{C}_{v}}=\dfrac{3}{2}$ which gives us the value of gamma as $\dfrac{5}{3}$.
Also, R = $\left( {{C}_{P}}-{{C}_{V}} \right)$ therefore, we can write the work done as ${{C}_{V}}\left( {{T}_{1}}-{{T}_{2}} \right)$ .
The work done is directly proportional to the quantity of gas in moles and the temperature difference.