Chemistry - Thermodynamics - I Question with Solution | TestHub

Given,

 $$\begin{gathered} {\mathrm{V}}_1=30 {\mathrm{dm}}^3 \\ {\mathrm{V}}_2=45 {\mathrm{dm}}^3 \\ \mathrm{P}=80 \mathrm{kPa} \end{gathered}$$

Using, first law of thermodynamics,
$\mathrm{\Delta U}=\mathrm{Q}+\mathrm{W}$

$\mathrm{Q}$ is the heat given or lost

$\mathrm{\Delta U}$ is the change in internal energy

$\mathrm{W}$ is the work done

$\mathrm{\Delta U}=0$ : Process is isothermal

$\mathrm{Q}=-\mathrm{W}$

$\mathrm{W}=-{\mathrm{P}}_{\mathrm{ext}} \mathrm{\Delta V}:$ Irreversible

$\mathrm{W}=-{\mathrm{P}}_{\mathrm{ext}} ({\mathrm{V}}_2-{\mathrm{V}}_1)$

$=-80\times {10}^3(45-30)\times {10}^{-3}$

$=-1200 \mathrm{J}$

Hence, the amount of heat transferred is $1200 \mathrm{J}$