Question

$1\mathrm{litre}$ of dry air at $\mathrm{STP}$ expands adiabatically to a volume of $3\mathrm{litres}$. If $\mathrm{\gamma }=1.4$, the work done by air is $\left(3^{1.4}=4.655\right)$ (take air to be an ideal gas)

• A. $48J$
• B. $100.8J$
• C. $90.5J$
• D. $60.7J$

Answer 1

The correct option is C

$90.5J$

Explanation for the correct answer:

Step 1: Given data

Dry air expands adiabatically at $\mathrm{STP}$.

i.e., ${\mathrm{P}}_1=1\mathrm{atm}$

$\Rightarrow {\mathrm{P}}_1=101.325\mathrm{K}\mathrm{Pa}$

$T_1=273\mathrm{K}$

Initial volume of Dry gas is $V_1=1\mathrm{L}$

$\mathit{\Rightarrow }{V}_1=1\times {10}^{-3}{\mathrm{m}}^3$

Volume of Dry gas after the expansion is $V_2=3\mathrm{L}$

$\mathit{\Rightarrow }{V}_2=3\times {10}^{-3}{\mathrm{m}}^3$

Specific heat capacity ratio of gas is $\mathrm{\gamma }=1.4$

Step 2: In the adiabatic process:

${\mathrm{P}}_1{\mathrm{V}}_1^{\mathrm{\gamma }}={\mathrm{P}}_2{\mathrm{V}}_2^{\mathrm{\gamma }}$

Substitute the given values to find ${\mathrm{P}}_2$.

$$\begin{gathered} \Rightarrow {\mathrm{P}}_2=\frac{{\mathrm{P}}_1·{\mathrm{V}}_1^{\mathrm{\gamma }}}{{\mathrm{V}}_2^{\mathrm{\gamma }}} \\ \Rightarrow {\mathrm{P}}_2=\frac{101.325\mathrm{K}\mathrm{Pa}·{\left(1\times {10}^{-3}{\mathrm{m}}^3\right)}^{1.4}}{{\left(3\times {10}^{-3}{\mathrm{m}}^3\right)}^{1.4}} \\ \Rightarrow {\mathrm{P}}_2=\left(\frac{101.325}{4.655}\right) \\ \Rightarrow {\mathrm{P}}_2=21.766\mathrm{K}\mathrm{Pa} \end{gathered}$$

Step 3: Work done in the adiabatic process:

$\mathrm{W}=-\left(\frac{{\mathrm{P}}_1{\mathrm{V}}_1-{\mathrm{P}}_2{\mathrm{V}}_2}{\mathrm{\gamma }-1}\right)$

Substitute the values to find the Work done in the adiabatic process.

$$\begin{gathered} \Rightarrow \mathrm{W}=-\left(\frac{\left(21.766\times {10}^3\mathrm{Pa}\times 3\times {10}^{-3}{\mathrm{m}}^3\right)-\left(101.325\times {10}^3\mathrm{Pa}\times 1\times {10}^{-3}{\mathrm{m}}^3\right)}{1.4-1}\right) \\ \Rightarrow \mathrm{W}=-\left(\frac{\left(65.29\right)-\left(101.325\right)}{0.4}\right) \\ \Rightarrow \mathrm{W}=-\left(\frac{-36.035}{0.4}\right) \\ \Rightarrow \mathrm{W}=-\left(-90.087\right) \\ \Rightarrow \mathrm{W}=90.087\mathrm{J} \end{gathered}$$

Hence, option C is correct. The work done by the dry air is $\mathrm{W}=90.087\mathrm{J}$.