Question

Calculate the maximum work done (in J) in expanding $16$ g of oxygen at $300$ K and occupying a volume of $5{\text{dm}}^3$ isothermally until the volume becomes $25{\text{dm}}^3$?

• A. $+2.11\times {10}^3$
• B. $+2.01\times {10}^3$
• C. $-2.01\times {10}^3$
• D. $-2.11\times {10}^3$

Answer 1

The correct option is C $-2.01\times {10}^3$

Maximum work is obtained in isothermal reversible process.
The expression for the maximum work is as follows:

$w=-2.303\text{nRT log}\frac{V_2}{V_1}$
The number of moles of oxygen is $N_{O_2}=\frac{\text{Weight}}{\text{Molecular weight}}=\frac{16}{32}=0.5$

The temperature is $T$ $=300$ K

The initial volume is $V_1$ $=5$ dm${}^3$

The final volume is $V_2$ $=25$ dm${}^3$

Substituting values in the above reaction, we get

$w=-2.303\times \frac{16}{32}\times 8.314\times 300\times \text{log}\frac{25}{5}=-2.01\times {10}^3$ J

Hence,option C is correct.