Chemistry - Thermodynamics - I Question with Solution | TestHub

Molecular weight of nitrous oxide $\left({\mathrm{N}}_2\mathrm{O}\right)=44\mathrm{g}/\mathrm{mol}$ Number of moles of ${\mathrm{N}}_2\mathrm{O}=\frac{2.2}{44}0.05\mathrm{mol}$ Specific heat capacity of ${\mathrm{N}}_2\mathrm{O}=880\mathrm{J}/\mathrm{g}.\mathrm{K}$ $\therefore$ Heat release on cooling from $310\mathrm{K}$ to $270\mathrm{K}$ $q=-2.2\times 880\times 40\mathrm{K}$ $q=-77.44\mathrm{KJ}$ Negative sign indicates heat released. $\therefore$ Work done by surrounding on system to compares the gas from $217.1\mathrm{ml}$ to $167.75\mathrm{ml}$ at constant pressure. It means work done by surrounding is irreversible. $\begin{array}{rl}& \therefore W=P\times \Delta V\\ & =1.01\times 10^5\times (167.75217.1\mathrm{ml}l)\\ & =1.01\times 10^5\times 0.049J\\ & W=4.95k.J\end{array}$ Using 1st Law of thermodynamics $\begin{array}{rl}& \Delta U=q+W\\ & \Delta U=-77.44KJ+4.95KJ\\ & \Delta U=-72.49KJ\end{array}$.
$\therefore x=72.$