Calculation the maximum work done in expanding 16 g of oxygen at 300 K and occupying a volume of 5 dm3 isothermally until volume becomes 25 dm3 .

Question

Calculation the maximum work done in expanding 16 g of oxygen at 300 K and occupying a volume of 5 dm3 isothermally until volume becomes 25 dm3 . (A)  - 2.01 × 103 J  (B)  + 2.01 × 103 J  (C)  2.01 J  (D)  2.01 × 103 kJ

Tardigrade Admin · Accepted Answer

Word done for isothermal expansion, W=-2.303 nRT log(Vf/Vi) =-2.303×(16/32)×8.314×300 log ((25/5)) =-2007.49 J =-2.01×103 J

Q. Calculation the maximum work done in expanding $16 g$ of oxygen at $300 K$ and occupying a volume of $5 d m^{3}$ isothermally until volume becomes $25 d m^{3}$ .

$- 2.01 \times 1 0^{3} J$

$+ 2.01 \times 1 0^{3} J$

$2.01 J$

$2.01 \times 1 0^{3} k J$

Word done for isothermal expansion,
$W = - 2.303 n RTl o g \frac{V _{f}}{V _{i}}$
$= - 2.303 \times \frac{16}{32} \times 8.314 \times 300 lo g (\frac{25}{5})$
$= - 2007.49 J$
$= - 2.01 \times 1 0^{3} J$

Q. Calculation the maximum work done in expanding 16g of oxygen at 300K and occupying a volume of 5dm3 isothermally until volume becomes 25dm3 .

−2.01×103J

+2.01×103J

2.01J

2.01×103kJ