Question

Initially, \(46.0\;{\rm{g}}\) oxygen is at a pressure of 1.00 atm and a temperature of \(400\;{\rm{K}}\). It expands adiabatically and reversibly until the pressure is reduced to 0.60atm, and it is then compressed isothermally and reversibly until the volume returns to its original value. Calculate the final pressure and temperature of the oxygen, the work done and heat added to the oxygen in this process, and the energy change dU. Take \({c_2}\left( {{{\rm{O}}_2}} \right) = 29.4\;{\rm{J}}\;{{\rm{K}}^{ - 1}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}\).

Short answer

The final pressure and temperature of the oxygen, the work done and heat added to the oxygen in this process, and the energy change are\(T = 346.63{\rm{K}}\),\({\rm{w}} = - 3.17{\rm{kJ}}\),\({\rm{q}} = 1.54\;{\rm{kJ}}\)and\(\Delta U = - 1.63\;{\rm{kJ}}\).

Step-by-step solution

Given data

The value of initial temperature, mass andinitial pressure:

\(\begin{array}{c}\;{\rm{T}} = 400{\rm{K}}\\mass = 46g\\p = 1.00atm\end{array}\)

The final pressure is\(0.6atm\)

Concept of adiabatical process and Isothermal process 

Adiabatical process: This is a process in which there is no heat transfer from the system to the surroundings or vice versa.

\(q = 0\)

This means that the internal energy change of the system is only dependent on the work done:

\(\Delta U = {w_{ad}}\)

Isothermal process:

This is a process in which the temperature stays constant.

Since there is no change in temperature, the change in internal energy will be\(0\).

Therefore, all of the work done will be released as heat or vice versa.

Calculate initial volume

The value of given mass and molar mass is given as:

\(\begin{array}{c}m = 46.0\;{\rm{g}}\\M = 32.00{\rm{gmo}}{{\rm{l}}^{ - 1}}\end{array}\)

The number of moles is given as:

\(\begin{array}{l}n = \frac{{m\left( {{{\rm{O}}_2}} \right)}}{{M\left( {{{\rm{O}}_2}} \right)}}\\n = 1.45\;{\rm{mol}}\end{array}\)

The initial volume is calculated byideal gas equation.

\(\begin{array}{c}pV = nRT\\V = \frac{{1.45 \times 0.08206 \times 400}}{1}\\{V_1} = 47.6\;{\rm{L}}\end{array}\)

Calculate final volume

Determine, \({ \vee _2}\):

Since given process is an adiabatical expansion, the relation between the pressure and volume is, \({p_1}V_1^\gamma = {p_2}V_2^\gamma \).

Where \(\gamma \) is the relation between the molar heat capacities.

\(\gamma = \frac{{{c_p}}}{{{c_v}}}\) ..…. (1)

For calculation of \(\gamma \) first determine the value of \({c_v}\) by the given formula:

\(\begin{array}{c}{c_p} = {c_v} + R\\{c_v} = {c_P} - R\\{c_v} = 29.4 - 8.314\\ = 21.1\end{array}\)

Put the value of \({c_p}\)and \({c_v}\) in equation (1).

\(\begin{array}{l}\gamma = \frac{{29.4}}{{21.1}}\\\gamma = 1.39\end{array}\)

Now calculate \({ \vee _2}\):

\(\begin{array}{c}{p_1}V_1^\gamma = {p_2}V_2^\gamma \quad {/^{\frac{1}{\gamma }}}\\p_1^{\frac{1}{\gamma }}{V_1} = p_2^{\frac{1}{\gamma }}{V_2}\\{V_2} = {\left( {\frac{{{p_1}}}{{{p_2}}}} \right)^{\frac{1}{\gamma }}} \times {V_1}\\{V_2} = {\left( {\frac{{1.00}}{{0.60}}} \right)^{\frac{1}{{1.33}}}} \times 47.6\end{array}\)

So, the final volume is \({V_2} = 68.74\;{\rm{L}}\).

Calculate temperature 

Now that we know the value of the volume we can go ahead and use the ideal gas law to calculate the temperature at the and of the adiabatical process.

\(\begin{array}{c}pV = nFT\\T = \frac{{pV}}{{nR}}\\T = \frac{{0.60 \times 68.74}}{{1.45 \times 0.08206}}\\{T_2} = 346.63\;{\rm{K}}\end{array}\)

Since the isothermal compression follows immediately after the adiabatical expansion.

\({T_2} = {T_3}\)

Therefore, the final temperature of the process will be \(346.63\;{\rm{K}}\).

Calculate work done in adiabatic processes

The work done is calculated as:

\(\begin{array}{c}w = n{c_v}\Delta T\\w = 1.45 \times 21.1 \times (346.63 - 400)\\{w_{ad}} = - 1.63\;{\rm{kJ}}\end{array}\)

Calculate change in internal energy

The work is equal to the change in internal energy in adiabatic processes.

\(\Delta {U_{ad}} = - 1.63\;{\rm{kJ}}\)

Calculate pressure 

Put the value all data to calculate the pressure from the ideal gas law:

\(\begin{array}{l}p = \frac{{nFT}}{V}\\p = \frac{{1.45 \times 0.08206 \times 346.63}}{{47.6}}\\{p_{{\rm{final }}}} = 0.87atm\end{array}\)

Calculate work done in isothermal process 

The work done in the isothermal process is calculated as:

\(\begin{array}{l}w = - nFT\ln \frac{{{V_2}}}{{{V_3}}}\\w = - 1.45 \times 8.314 \times 346.63 \times \ln \frac{{68.74}}{{47.6}}\\{w_{is}} = - 1.54\;{\rm{kJ}}\end{array}\)

Since we determined the relation:

\(q = - w\)

So, \({q_{is}} = 1.54\;{\rm{kJ}}\).

Calculate overall work done

Calculate the overall work done:

\(\begin{array}{l}w = {w_{ad}} + {w_{is}}\\w = - 1.63 - 1.54\\w = - 3.17\;{\rm{kJ}}\end{array}\)

Calculate overall heat

The overall heat is calculated as:

\(\begin{array}{l}q = {q_{kd}} + {q_{is}}\\w = 0 + 1.54\\q = 1.54kJ\end{array}\)

Calculate overall change in internal energy

The overall internal energy change is calculated as:

\(\begin{array}{l}\Delta U = \Delta {U_{ad}} + \Delta {U_{is}}\\\Delta U = - 1.63 + 0\\\Delta U = - 1.63\;{\rm{kJ}}\end{array}\)