Solution Found!
0.10 mol of gas undergoes the process 1 ? 2 shown in
Chapter 16, Problem 59P(choose chapter or problem)
\(0.10 \mathrm{~mol}\) of gas undergoes the process \(1 \rightarrow 2\) shown in Figure P16.59.
a. What are temperatures \(T_{1}\) and \(T_{2}\left(\right.\) in \(\left.{ }^{\circ} \mathrm{C}\right)\)?
b. What type of process is this?
c. The gas undergoes an isothermal compression from point 2 until the volume is restored to the value it had at point 1 . What is the final pressure of the gas?
Questions & Answers
QUESTION:
\(0.10 \mathrm{~mol}\) of gas undergoes the process \(1 \rightarrow 2\) shown in Figure P16.59.
a. What are temperatures \(T_{1}\) and \(T_{2}\left(\right.\) in \(\left.{ }^{\circ} \mathrm{C}\right)\)?
b. What type of process is this?
c. The gas undergoes an isothermal compression from point 2 until the volume is restored to the value it had at point 1 . What is the final pressure of the gas?
ANSWER:
Step 1 of 3
a.)
We have to find the temperatures \(T_{1}\) and \(T_{2}\) of \(0.1 \mathrm{~mol}\) of a gas which undergoes the process \(1 \rightarrow 2\) shown in the given figure.
The gas undergoes an isobaric process (constant pressure) as the initial and final pressure ( \(p_{1}=p_{2}=1 \mathrm{~atm}\) ) are the same which can be seen from the figure.
The temperatures \(T_{1}\) and \(T_{2}\) of the gas can be found using the ideal gas equation.
\(T_{1}=\frac{p_{1} V_{1}}{n R}\)
Where,
\(p_{1}=1 \mathrm{~atm}=101300 \mathrm{~Pa}\)
\(V_{1}=1000 \mathrm{~cm}^{3}=1000 \times 10^{-6} \mathrm{~m}^{3} \text { (from figure) }\)
\(n=0.1\)
\(R=8.31 \mathrm{~J} / \mathrm{mol} \mathrm{K}\)
Thus,
\(T_{1}=\frac{\left(101300 \times 1000 \times 10^{-6}\right)}{(0.6 \times 8.31)}\)
\(=122 \mathrm{~K}\)
\(=-152{ }^{\circ} \mathrm{C}\)
\(T_{2}=\frac{p_{2} V_{2}}{n R}\)
Where,
\(p_{2}=1 \mathrm{~atm}=101300 \mathrm{~Pa}\)
\(V_{2}=3000 \mathrm{~cm}^{3}=3000 \times 10^{-6} \mathrm{~m}^{3}(\text { from figure })\)
\(n=0.1\)
\(R=8.31 \mathrm{~J} / \mathrm{mol} \mathrm{K}\)
Thus,
\(T_{2}=\frac{\left(101300 \times 3000 \times 10^{-6}\right)}{(0.6 \times 8.31)}\)
\(=366 \mathrm{~K}\)
\(=93{ }^{\circ} \mathrm{C}\)
Therefore, the temperatures \(T_{1}\) and \(T_{2}\) are \(-152^{\circ} \mathrm{C}\) and \(93^{\circ} \mathrm{C}\) respectively.