Question

During an adiabatic compression, $830\mathrm{J}$ of work is done on $2$ moles of a diatomic ideal gas to reduce its volume by $50\%$. The change in its temperature is nearly: [$\mathrm{R}=8.3{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1}$]

• A. $14\mathrm{K}$
• B. $20\mathrm{K}$
• C. $24\mathrm{K}$
• D. $40\mathrm{K}$

Answer 1

The correct option is B

$20\mathrm{K}$

Explanation:

Step 1: Analyzing the given data

$\mathrm{Work}\mathrm{done}\left(\mathrm{W}\right)=830\mathrm{J}$,

number of moles$\left(\mathrm{n}\right)=2$

$\mathrm{R}=8.3{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1}$;

Gas is diatomic, so the value of the adiabatic constant $\mathrm{\gamma }=1.4$

Step 2: Calculating the change in temperature

For the adiabatic process the expression for work done is given as:

$\mathrm{W}=\frac{\mathrm{PV}}{\mathrm{\gamma }-1}$

By ideal gas equation, we know that, $\mathrm{PV}=\mathrm{nR\Delta T}$

The equation for the work done can be re-written as: $\mathrm{W}=\frac{\mathrm{nR\Delta T}}{\mathrm{\gamma }-1}$

$$\begin{gathered} \mathrm{W}=\frac{2\mathrm{mole}\times 8.3{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1}\times \mathrm{\Delta T}}{1.4-1} \\ 830\mathrm{J}=\frac{16.6{\mathrm{JK}}^{-1}\times \mathrm{\Delta T}}{0.4} \\ \mathrm{\Delta T}=20\mathrm{K} \end{gathered}$$

So, option(B) is correct.