Question

$16$gm of $O_2$ expand at constant pressure and $273.15$K to occupy double its original volume. The work done during the process will be:

• A. $w=-260$cal
• B. $w=-180$cal
• C. $w=-130$cal
• D. $w=-273.15$cal

Answer 1

The correct option is D $w=-273.15$cal

$Workdone=-P_{ext}dV$

$\Rightarrow -P_{ext}\left(V_i-V_i\right)$

$\Rightarrow -P_{ext}\left(2V_i-V_i\right)$ $V_t=2V_i$

$\Rightarrow -P_{ext}{V}_i$

$\Rightarrow -nRT$

$W=-\frac{16}{32}\times 2Cal{K}^{-1}mo{l}^{-1}\times 273.15$

$=-2\times 273.15Cal$

$W=-273.15Cal$

Hence, the correct option is $\text{D}$