Question

In an isothermal reversible expansion, if the volume of $96\text{g}$ of oxygen at ${27}^{\circ }\text{C}$ is increased from $70\text{litres}$ to $140\text{litres}$, then the work done by the gas will be

• A. $300R{\log }_{10}2$
• B. $2.3\times 900R{\log }_{10}2$
• C. $600R{\log }_{10}2$
• D. $2.3\times 300R{\log }_{10}2$

Answer 1

The correct option is B $2.3\times 900R{\log }_{10}2$

Given,
Initial volume of oxygen $\left(V_i\right)=70\text{liters}$
Final volume of oxygen , $\left(V_f\right)=140\text{liters}$
Mass of the oxygen $\left(m\right)=96\text{g}$
Temperature is maintained at $T={27}^{\circ }\text{C}$
we know that,
Molecular mass of oxygen $\left(M\right)=32\text{g}$
Let us assume that $\mu$ is the number of moles of oxygen
$\because$ we know that, work done by gas in an isothermal process is given by:
$W=\mu RT{\text{log}}_e\frac{V_f}{V_i}$
By substituting the values from the data given in the question, we get
$W=\left(\frac{m}{M}\right)RT{\log }_e\frac{V_2}{V_1}=2.3\times \frac{m}{M}\times RT{\log }_{10}\frac{V_f}{V_i}$
$=2.3\times \frac{96}{32}\times R\times 300\times {\log }_{10}\frac{140}{70}=2.3\times 900R{\log }_{10}2$
Thus, option (b) is the correct answer.