Chemistry - Thermodynamics - 1 Question with Solution | TestHub

Here,$\mathrm{\Delta }\mathrm{V}=200-400=-200\mathrm{ }\text{c}{\text{m}}^3$
A we know$1\text{Litre}=1000{\text{ cm}}^{\text{3}}$
$\Rightarrow \text{ΔV}=-0.2\text{Litre}$.
External Pressure$\left(\mathrm{P}\right)=0.50$then
Work done$=-\mathrm{P}\mathrm{\Delta }\mathrm{V}=+0.50\left(0.2\right)$
$=+0.1\text{atm}\text{Litre}$
and$\text{1}\text{litre-atmosphere}\text{=}\text{101}\text{.3}\text{Joule}$
$\therefore \mathrm{W}=+10.13$Joule
According to First law of thermodynamics
$\mathrm{q}=\mathrm{\Delta }\mathrm{E}-\mathrm{W}$
then$\mathrm{\Delta }\mathrm{E}=\mathrm{q}+\mathrm{W}$
Given$\mathrm{q}=-8.00\mathrm{ }\mathrm{J}$
$\Rightarrow \text{ΔE}=-8+10.13=2.13\text{Joule}$