Question

An ideal gas expands from a volume of $6d{m}^3$ to $16d{m}^3$ against the constant external pressure of $2.026\times {10}^{5}N{m}^{-}2$. Find enthalpy change if $\Delta u$ is $418J$.

Answer 1

Given that:

$\begin{array}{c}{V}_1=6d{m}^3=6\times {\left(0.1m\right)}^3=6\times {10}^{-3}{m}^3\\ V_2=16d{m}^3=16\times {10}^{-3}{m}^3\\ P_{ext}=2.026\times {10}^{5}N{m}^{-2}\\ \Delta u=418J\\ work,done=-P\Delta V\\ w=-2.026\times {10}^5\left(V_2-V_1\right)\\ w=-2.026\times {10}^5\left(6\times {10}^{-3}-16\times {10}^{-3}\right)Nm\\ w=-2.026\times {10}^5\left(-10\times {10}^{-3}\right)Joule\\ w=-2.026\times {10}^{3}Joule\\ we,have\\ \Delta H=\Delta u+P\Delta V\\ \Delta H=418+2.026\times {10}^{3}J\\ \Delta H=2444J\end{array}$