Question

For nitrogen gas the values of ${\mathit{C}}_{\mathbf{v}}$, and ${\mathit{C}}_{\mathbf{p}}$ at 25°C are $\mathbf{20}\mathbf{.}\mathbf{8}\mathit{J}{\mathit{K}}^{\mathbf{-}\mathbf{1}}\mathit{m}\mathit{o}{\mathit{l}}^{\mathbf{-}\mathbf{1}}\mathit{a}\mathit{n}\mathit{d}\mathbf{29}\mathbf{.}\mathbf{1}\mathit{J}{\mathit{K}}^{\mathbf{-}\mathbf{1}}\mathit{m}\mathit{o}{\mathit{l}}^{\mathbf{-}\mathbf{1}}$, respectively. When a sample of nitrogen is heated at constant pressure, what fraction of the energy is used to increase the internal energy of the gas? How is the remainder of the energy used? How much energy is required to raise the temperature of 100.0 g N from 25.0°C to 85.0°C in a vessel having a constant volume?

Short answer

Theenergy is required to raisethe temperature of 100.0 g N from 25.0°C to 85.0°C in a vessel having a constant volume is $\mathbf{4}\mathbf{.}\mathbf{46}\mathbf{}\mathit{K}\mathit{J}$.

Step-by-step solution

Introduction to the Concept

The change in heat for a system to modify its internal energy is denoted as $\mathit{\Delta }\mathit{E}\mathbf{}\mathit{o}\mathit{r}\mathbf{}{\mathit{q}}_{\mathbf{v}}$ at constant volume. The change in heat for a system to alter its enthalpy at constant pressure is denoted as$\mathit{\Delta }\mathit{H}\mathbf{}\mathit{o}\mathit{r}\mathbf{}{\mathit{q}}_{\mathbf{\rho }}$.

According to the first rule of thermodynamics,

$\Delta E=q+w$

Where,

$q=heat$

$w=work$

$\Delta E=\text{Internal Energy}$

Step 2: Determination of fraction

From the given data,

$C_v=20.8\text{J/Kmol}$

$C_{\rho }=29.1\text{J/Kmol}$

Mass of $N_2=100.0\text{g}$

Initial Temperature$=25.0°C$

Final Temperature$=85.0°C$

Let’s calculate fraction,

$\mathit{F}\mathit{r}\mathit{a}\mathit{c}\mathit{t}\mathit{i}\mathit{o}\mathit{n}\mathbf{=}\frac{{\mathbf{C}}_{\mathbf{v}}}{{\mathbf{C}}_{\mathbf{p}}}$

So, Fraction$=\frac{20.8}{29.1}$

$=0.714$

So,$0.714$fraction is used to increase the internal energy of the system.

The remaining energy is used to do pressure-volume work.

Step 3: Determination of Energy

Let’s compute the energy at constant V,

$\mathit{\Delta }\mathit{E}\mathbf{=}\mathit{n}\mathbf{\times }{\mathit{C}}_{\mathbf{v}}\mathbf{\times }\mathit{\Delta }\mathit{T}$

$\Delta E=\frac{100.0\text{g}}{28.0g/mol}\times 20.8J/Kmol\times \left(85-25\right)°C$

$=4457J$

$\Delta E=4.46\text{K}J$

Therefore the internal energy is $\mathbf{4}\mathbf{.}\mathbf{46}\mathbf{}\mathit{K}\mathit{J}$.