Question: (III) The PV diagram in Fig. 15–23 shows two possible states of a system containing 1.75 moles of a monatomic ideal gas. \(\left( {{P_1} = {P_2} = {\bf{425}}\;{{\bf{N}} \mathord{\left/{\vphantom {{\bf{N}} {{{\bf{m}}^{\bf{2}}}}}} \right.} {{{\bf{m}}^{\bf{2}}}}},\;{V_1} = {\bf{2}}{\bf{.00}}\;{{\bf{m}}^{\bf{3}}},\;{V_2} = {\bf{8}}{\bf{.00}}\;{{\bf{m}}^{\bf{3}}}.} \right)\) (a) Draw the process which depicts an isobaric expansion from state 1 to state 2, and label this process A. (b) Find the work done by the gas and the change in internal energy of the gas in process A. (c) Draw the two-step process which depicts an isothermal expansion from state 1 to the volume \({V_2}\), followed by an isovolumetric increase in temperature to state 2, and label this process B. (d) Find the change in internal energy of the gas for the two-step process B.

An isobaric process is a process that is carried out at constant pressure. In this process, the reaction’s heat is equal to the change in the enthalpy of the system.

Given data:

The number of moles of a monoatomic gas is\(n = 1.75\;{\rm{moles}}\).

The pressure is \(P = {P_1} = {P_2} = 425\;{{\rm{N}} \mathord{\left/{\vphantom {{\rm{N}} {{{\rm{m}}^{\rm{2}}}}}} \right.} {{{\rm{m}}^{\rm{2}}}}}\).

The initial volume is \({V_1} = 2.00\;{{\rm{m}}^{\rm{3}}}\).

The final volume is \({V_2} = 8.00\;{{\rm{m}}^{\rm{3}}}\).

(a)

The following graph shows the process of isobaric expansion from state 1 to 2.

(b)

The work done by the gas when its volume changes from \({V_1}\) to \({V_2}\) can be calculated as:

\(\begin{aligned}{c}W &= P\left( {{V_2} - {V_1}} \right)\\W &= \left( {425\;{{\rm{N}} \mathord{\left/{\vphantom {{\rm{N}} {{{\rm{m}}^{\rm{2}}}}}} \right.} {{{\rm{m}}^{\rm{2}}}}}} \right)\left( {\left( {8.00\;{{\rm{m}}^{\rm{3}}}} \right) - \left( {2.00\;{{\rm{m}}^{\rm{3}}}} \right)} \right)\\W &= 2550\;{\rm{J}}\end{aligned}\)

Thus, the work done is \(2550\;{\rm{J}}\).

The expression for the change in internal energy of gas when temperature changes from \({T_1}\) to \({T_2}\) is as follows:

\(\Delta U = \frac{3}{2}nR\left( {{T_2} - {T_1}} \right)\)

From ideal gas equation, \(nR{T_1} = P{V_1}\) and \(nR{T_2} = P{V_2}\). Thus, the expression for the change in internal energy becomes:

\(\begin{aligned}{c}\Delta U &= \frac{3}{2}\left( {P{V_1} - P{V_2}} \right)\\\Delta U &= \frac{3}{2}P\left( {{V_2} - {V_1}} \right)\\\Delta U &= \frac{3}{2}W\end{aligned}\)

Substitute the values in the above equation.

\(\begin{aligned}{c}\Delta U &= \frac{3}{2}\left( {2550\;{\rm{J}}} \right)\\\Delta U &= 3830\;{\rm{J}}\end{aligned}\)

Thus, the change in internal energy is \(3830\;{\rm{J}}\).

(c)

The below PV diagram shows the isothermal expansion from state 1 to 3 and isovolumetric process from state 3 to 2.

(d)

The change in internal energy depends on temperature change. Here, the temperature change is same for process A and B. Therefore, the change in internal energy in process B will be \(3830\;{\rm{J}}\).