Question

Consider a rigid, insulated box containing $\mathbf{0}\mathbf{.}\mathbf{400}\mathbf{}\mathit{m}\mathit{o}\mathit{l}\mathit{e}\mathbf{}\mathit{o}\mathit{f}\mathbf{}\mathit{H}\mathit{e}\left(g\right)\mathbf{}\mathit{a}\mathit{t}\mathbf{}\mathbf{20}\mathbf{.}\mathbf{0}\mathbf{°}\mathit{C}$and 1.00 atm in one compartment and $\mathbf{0}\mathbf{.}\mathbf{600}\mathbf{}\mathit{m}\mathit{o}\mathit{l}\mathit{e}\mathbf{}\mathit{o}\mathit{f}\mathbf{}{\mathit{N}}_{\mathbf{2}}\left(g\right)\mathbf{}\mathit{a}\mathit{t}\mathbf{}\mathbf{100}\mathbf{.}\mathbf{0}\mathbf{°}\mathit{C}$and 2.00 atm in the other compartment. These compartments are connected by a partition that transmits heat. What is the final temperature in the box at thermal equilibrium?

$\left[ForHe\left(g\right),C_V=12.5J{K}^{-1}mo{l}^{-1};for{N}_2\left(g\right),C_v=20.7J{K}^{-1}mo{l}^{-1}.\right]$

Short answer

The final temperaturein the box at thermal equilibrium is${\mathbf{77}}^{\mathbf{o}}\mathit{C}$

Step-by-step solution

Step 1: Introduction to the Concept

Heat is exchanged between the system (reactants and products) and the environment in general during chemical reactions. Underconstant volume circumstances, the heat (q) absorbed or evolved is given as:

${\mathit{q}}_{\mathbf{v}}\mathbf{=}\mathit{n}\mathbf{\times }{\mathit{C}}_{\mathbf{v}}\mathbf{\times }\mathit{\Delta }\mathit{T}\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{-}\left(1\right)$

Where, $\mathit{n}\mathit{a}\mathit{n}\mathit{d}{\mathit{C}}_{\mathbf{V}}$ are the number of moles and specific heat capacity of the material at constant volume respectively.

$\mathit{\Delta }\mathit{T}\mathbf{=}\mathit{C}\mathit{h}\mathit{a}\mathit{n}\mathit{g}\mathit{e}\mathit{i}\mathit{n}\mathit{t}\mathit{e}\mathit{m}\mathit{p}\mathit{e}\mathit{r}\mathit{a}\mathit{t}\mathit{u}\mathit{r}\mathit{e}\mathbf{=}{\mathit{T}}_{\mathbf{final}}\mathbf{-}{\mathbf{T}}_{\mathbf{initial}}$

Step 2: Determination of heat (q) lost by N2 gas

From the given,

Number of He Moles$=0.400$

Number of $N_2$Moles $=0.600$

Temperature:

$He=T_{He}=20.00°C$

$N_2=T_{N_2}=100.00°C$

Specific heat:

$C_{v\left(He\right)}=12.5\text{J/mol}K$

$C_{v\left(N_2\right)}=20.7 J/molK$

From the equation (1), heat lost is taken as negative.

$q_{N_2}=-n_{N_2}\times C_{v\left(N_2\right)}\times \left(T_f-T_{N_2}\right)$

$q_{N_2}=-12.42\times \left(T_f-373\right)\text{J}$

Determination of heat (q) gained by  gas

$q_{He}=-n_{He}\times C_{V\left(He\right)}\times \left(T_f-T_{He}\right)$

$q_{He}=5\times \left(T_f-293\right)\text{J}$

Step 4: Determination of final temperature

Thus, equate both,

$\mathit{H}\mathit{e}\mathit{a}\mathit{t}\mathbf{}\mathit{l}\mathit{o}\mathit{s}\mathit{t}\mathbf{=}\mathit{h}\mathit{e}\mathit{a}\mathit{t}\mathbf{}\mathit{g}\mathit{a}\mathit{i}\mathit{n}\mathit{e}\mathit{d}$

${\mathit{q}}_{{\mathbf{N}}_{\mathbf{2}}}\mathbf{=}{\mathit{q}}_{\mathbf{H}\mathbf{e}}$

$-12.42\times \left(T_f-373\right)\text{J}=5\times \left(T_f-293\right)\text{J}$

$T_f=350\text{K}$

i.e$T_f=\left(350-273\right)K$

$T_f=77°C$

Therefore the final temperature is ${\mathbf{77}}^{\mathbf{o}}\mathit{C}$.