Question

A sample of$\mathbf{1}\mathbf{.}\mathbf{75}\mathit{m}\mathit{o}\mathit{l}\mathit{e}\mathit{s}$of ${\mathit{H}}_{\mathbf{2}}\mathbf{(}{\mathit{C}}_{\mathbf{v}}\mathbf{=}\mathbf{20}\mathbf{.}\mathbf{5}\mathit{J}{\mathit{K}}^{\mathbf{-}\mathbf{1}}\mathit{m}\mathit{o}{\mathit{l}}^{\mathbf{-}\mathbf{1}}\mathbf{)}$at${\mathbf{21}}^{\mathbf{o}}\mathit{C}$ and $\mathbf{1}\mathbf{.}\mathbf{50}\mathit{a}\mathit{t}\mathit{m}$undergoes a reversible adiabatic compression until the final pressure is$\mathbf{4}\mathbf{.}\mathbf{50}\mathit{a}\mathit{t}\mathit{m}$. Calculate the final volume of the gas sample and the work associated with this process. Assume that the gas behaves ideally

Short answer

The final volume for an adiabatic process is$12.8\text{\hspace{0.17em}L}$

The work associated for this process is$3.84\text{\hspace{0.17em}kJ}$

Step-by-step solution

Definition of adiabatic process

A process in which no heat transfer takes place is defined as adiabatic process

Calculations

Amount of heat required at constant temperature is:

$\begin{array}{l}{\text{C}}_{\text{p}}=28.82\text{\hspace{0.17em}J/K\hspace{0.17em}mol}\\ \gamma =\frac{{\text{C}}_{\text{p}}}{{\text{C}}_{\text{v}}}\\ \gamma =1.40\end{array}$

$\begin{array}{l}{\text{V}}_{\text{2}}=12.8\text{L}\\ {\text{T}}_{\text{2}}=401\text{K}\end{array}$

The work associated is:

$\begin{array}{l}\text{w}={\text{nC}}_{\text{v}}\text{ΔT}\\ \text{w}=\left(1.75\text{\hspace{0.17em}mol}\right)\left(20.5\text{\hspace{0.17em}J/K\hspace{0.17em}mol}\right)(401\text{\hspace{0.17em}K}-294\text{\hspace{0.17em}K})\\ \text{w}=3838.625\text{\hspace{0.17em}J}\end{array}$

Converting the value in joule to kilo joule,

$\begin{array}{l}1\text{\hspace{0.17em}kJ}=1000\text{\hspace{0.17em}J}\\ \text{w}=3838.625\text{\hspace{0.17em}J}\times \frac{1\text{\hspace{0.17em}kJ}}{1000\text{\hspace{0.17em}J}}\\ \text{w}=3.84\text{\hspace{0.17em}kJ}\end{array}$