Solution Found!
0.0050 mol of gas undergoes the process 1 ? 2 ? 3 shown in
Chapter 16, Problem 60P(choose chapter or problem)
\(0.0050$ mol of gas undergoes the process \(1 \rightarrow 2 \rightarrow 3\) shown in Figure P16.60. What are (a) temperature \(T_{1}\), (b) pressure \(p_{2}\), and (c) volume \(V_{3} ?\)
Equation Transcription:
Text Transcription:
0.0050 mol
1 rightarrow 2 rightarrow 3
T_1
p_2
V_3
Questions & Answers
QUESTION:
\(0.0050$ mol of gas undergoes the process \(1 \rightarrow 2 \rightarrow 3\) shown in Figure P16.60. What are (a) temperature \(T_{1}\), (b) pressure \(p_{2}\), and (c) volume \(V_{3} ?\)
Equation Transcription:
Text Transcription:
0.0050 mol
1 rightarrow 2 rightarrow 3
T_1
p_2
V_3
ANSWER:
Step 1 of 3
a.)
We have to find the temperature \(\left(T_{1}\right)\) of \(.0050 \mathrm{~mol}\) of gas which undergoes the process \(1 \rightarrow 2 \rightarrow 3\) as shown in figure.
The volume and pressure of the gas in state 1 from the figure is found to be equal to \(100 \mathrm{~cm}^{3}\) and \(1 \mathrm{~atm}\) respectively.
The temperature of the gas in state 1 can be found using the ideal gas law.
\(T_{1}=\frac{p_{1} V_{1}}{n R}\)
where,
\(p_{1}=1 \mathrm{~atm}=101300 \mathrm{~Pa}\)
\(V_{1}=100 \mathrm{~cm}^{3}=100 \times 10^{-6} \mathrm{~m}^{3}\)
\(R=8.31 \mathrm{~J} / \mathrm{mol} \mathrm{K}\)
\(n=0.0050\)
Thus,
\(T_{1}=\frac{\left(101300 \times 100 \times 10^{-6}\right)}{(0.0050 \times 8.31)}\)
\(=244 \mathrm{~K}\)
\(=-29{ }^{\circ} \mathrm{C}\)
Therefore, the temperature \(T_{1}\) is \(-29^{\circ} \mathrm{C}\).