Question

Question: (II) 8.5 moles of an ideal monatomic gas expand adiabatically, performing 8300 J of work in the process. What is the change in temperature of the gas during this expansion?

Short answer

The change in temperature of the gas during this expansion is \( - 78.3\;{\rm{K}}\).

Step-by-step solution

Understanding of adiabatic process

An adiabatic process may be defined as the process that happens without any heat exchange between the surroundings and the system. In this process, the work done is due to a change in its internal energy.

Given information

Given data:

The number of moles is\(n = 8.5\;{\rm{moles}}\).

The work done by the gas is \(W = 8300\;{\rm{J}}\).

Evaluation of the change in temperature of the gas during this expansion process

No heat transfer occurs in the adiabatic process between the surroundings and the system. Therefore, heat flows out or into the gas is zero. That is \(Q = 0\).

According to the first law of thermodynamic, the change in the internal energy of the system will be:

\(\begin{aligned}{c}\Delta U &= Q - W\\\Delta U &= 0 - W\\\Delta U &= - W\end{aligned}\) … (i)

The expression for the change in the internal energy for a monoatomic gas is as follows:

\(\Delta U = \frac{3}{2}nR\Delta T\) … (ii)

Here, \(R\) is the universal gas constant and its value is \(8.314\;{{\rm{J}} \mathord{\left/{\vphantom {{\rm{J}} {{\rm{mol}} \cdot {\rm{K}}}}} \right.} {{\rm{mol}} \cdot {\rm{K}}}}\).

Now equate the equation (i) and (ii).

\(\begin{aligned}{c}\frac{3}{2}nR\Delta T &= - W\\\Delta T &= - \frac{2}{3}\frac{W}{{nR}}\end{aligned}\)

Substitute the values in the above equation.

\(\begin{aligned}{c}\Delta T &= - \left( {\frac{2}{3}} \right)\frac{{\left( {8300\;{\rm{J}}} \right)}}{{\left( {8.5\;{\rm{moles}}} \right)\left( {8.314\;{{\rm{J}} \mathord{\left/{\vphantom {{\rm{J}} {{\rm{mol}} \cdot {\rm{K}}}}} \right.} {{\rm{mol}} \cdot {\rm{K}}}}} \right)}}\\\Delta T &= - 78.3\;{\rm{K}}\end{aligned}\)

Thus, the change in temperature of the gas during this expansion is \( - 78.3\;{\rm{K}}\).