Question

16 g oxygen expands at STP isobarically to occupy double of its original volume. The work done during the process will be:

• A. 266 kcal
• B. 187.905 kcal
• C. 130 kcal
• D. 272.84 kcal

Answer 1

Answer: (B)

187.905 kcal

$32$ grams $O_2=1$ mole

$16$ grams $O_2=0.5$ mole

$16\times {10}^3$ gram $=16kg$ $=500$ moles of $O_2\left(n\right)$

We know that,

$W=-nPV$, Now we know, At $STP$, $T=273K$

$W=nRTIn\frac{V_2}{V_1}$

Here $n=$ number of moles $=\frac{16000}{32}=500R=1.986$ cal

$T=273K$

$\frac{V_2}{V_1}=2$

So, $W=500x1.986\times 273\times In2$

$=187905$ cal

$=187.905$ K.cal