Question

\[\text{2}\text{.8 g}\] of ${{\text{N}}_{\text{2}}}$ at \[\text{300K}\] and \[\text{20atm}\] was allowed to expand isothermally against a constant external pressure of\[\text{1 atm}\]. Calculate\[\text{ }\!\!\Delta\!\!\text{ U}\], q and W for the gas.

Answer 1

Hint: According to the first law of thermodynamic the internal energy, heat and work are related as $\text{ }\!\!\Delta\!\!\text{ U = q + w}$.here the nitrogen undergoes the irreversible isothermal expansion. Therefore the work associated with the expansion is $-{{\text{P}}^{\text{ext}}}\left( {{\text{V}}_{\text{2}}}-{{\text{V}}_{\text{1}}} \right)$.

Complete step by step solution:
The ${{\text{N}}_{\text{2}}}$ gas undergoes the isothermal expansion. The data given is as follows:
\[\begin{align}
& \text{Weight of }{{\text{N}}_{\text{2}}}\text{ gas = 2}\text{.8 g} \\
& \text{Absolute temperature = 300K} \\
& \text{Pressure }{{\text{P}}_{\text{1}}}\text{ = 20atm} \\
& \text{Pressure }{{\text{P}}_{\text{2}}}\text{ = 1 atm} \\
& \text{To find : } \\
& \text{1) }\!\!\Delta\!\!\text{ U} \\
& \text{2) q} \\
& \text{3) w} \\
\end{align}\]

1) Lets first find the \[\text{ }\!\!\Delta\!\!\text{ U}\]. The \[\text{ }\!\!\Delta\!\!\text{ U}\]is the energy associated with the substance which depends on the chemical nature as well as the temperature, pressure, and volume of the substance.
In an isothermal process, the temperature of the system remains constant throughout the process of expansion. Since for an ideal gas, the internal enthalpy U depends only on temperature, it follows that at constant temperature (isothermal process), the internal energy of the gas remains constant. This means that the \[\text{ }\!\!\Delta\!\!\text{ U= 0}\]

Therefore, the internal energy for isothermal expansion is zero.

2) Let’s calculate the work done by the gas during isothermal expansion. The work done for isothermal expansion is given as follows:
$\text{w= }-\text{ }\int\limits_{{{\text{V}}_{\text{1}}}}^{{{\text{V}}_{\text{2}}}}{{{\text{P}}^{\text{ext}}}\text{.dV =}}\text{ }-{{\text{P}}^{\text{ext}}}\left( {{\text{V}}_{\text{2}}}-{{\text{V}}_{\text{1}}} \right)$ (1)
Let’s first calculate the number of moles of nitrogen gas.
$\begin{align}
  & \text{No}\text{. of moles of }{{\text{N}}_{\text{2}}}\text{ = }\dfrac{\text{weight of }{{\text{N}}_{\text{2}}}\text{ }}{\text{Molecular weight of }{{\text{N}}_{\text{2}}}} \\
 & \therefore \text{ n = }\dfrac{2.8}{28}=\text{ 0}\text{.1} \\
\end{align}$
We have to find out the volume at the initial and final conditions. Thus let’s first calculate the volume of gas at the initial condition. Here, the pressure is\[{{\text{P}}_{\text{1}}}\text{ = 20atm}\]. Let’s apply the ideal gas equation to get the value of volume${{\text{V}}_{\text{1}}}$.
\[\begin{align}
  & {{\text{P}}_{\text{1}}}{{\text{V}}_{\text{1}}}\text{=nRT} \\
 & \Rightarrow {{\text{V}}_{\text{1}}}\text{ = }\dfrac{\text{nRT}}{{{\text{P}}_{\text{1}}}}=\text{ }\dfrac{\left( 0.1 \right)\left( \text{R} \right)\left( 300 \right)}{20}\text{ L } \\
 & \therefore \text{ }{{\text{V}}_{\text{1}}}\text{ = 1}\text{.5 R} \\
\end{align}\]
Let us find out the final volume ${{\text{V}}_{2}}$ for the pressure\[{{\text{P}}_{2}}\text{ = 1 atm}\]. Let’s apply the ideal gas equation to get the value of${{\text{V}}_{2}}$.
\[\begin{align}
  & {{\text{P}}_{2}}{{\text{V}}_{2}}\text{=nRT} \\
 & \Rightarrow {{\text{V}}_{2}}\text{ = }\dfrac{\text{nRT}}{{{\text{P}}_{2}}}=\text{ }\dfrac{\left( 0.1 \right)\left( \text{R} \right)\left( 300 \right)}{1}\text{ L } \\
 & \therefore \text{ }{{\text{V}}_{2}}\text{ = 30 R} \\
\end{align}\]
Let us substitute the values of ${{\text{V}}_{\text{1}}}$ and ${{\text{V}}_{2}}$ in equation (1). We get,
$\begin{align}
  & \text{ w = }-{{\text{P}}^{\text{ext}}}\left( {{\text{V}}_{\text{2}}}-{{\text{V}}_{\text{1}}} \right) \\
 & \Rightarrow \text{w = }-\text{1}\left( 30\text{R}-1.5\text{R} \right) \\
 & \Rightarrow \text{w = }-\text{ 28}\text{.5}\times \text{ 8}\text{.314} \\
 & \therefore \text{ w = }-\text{ 236}\text{.95 J} \\
\end{align}$

Therefore, work done is equal to the$-\text{ 236}\text{.95 J}$.

3) Let's find out the heat of the system. According to the first law of thermodynamics,$\text{ }\!\!\Delta\!\!\text{ U = q + w}$ where q is the heat of the system and w is the work done.
We know that for isothermal expansion the internal energy is equal to zero. Thus the first law of thermodynamics is modified as,
$\begin{align}
& \text{ }\!\!\Delta\!\!\text{ U = q + w} \\
& \text{ 0 = q + w} \\
& \therefore \text{ q = }-\text{w} \\
\end{align}$
Let us substitute the value of work from above to find the heat. We get,
$\begin{align}
  & \text{q = }-\text{w} \\
 & \text{q = }-(-236.95\text{ J)} \\
 & \therefore \text{ q = }236.95\text{ J} \\
\end{align}$

Therefore the heat involved in the isothermal expansion of nitrogen gas is $236.95\text{ J}$.

Note:The gas undergoes the isothermal expansion in presence of external pressure thus the formula we have applied the formula used for irreversible isothermal expansion as the changes cannot be undone. For example, the balloon burst into the atmosphere.