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Solved: 0.10 mol of argon gas is admitted to an evacuated
Chapter 16, Problem 28E(choose chapter or problem)
\(0.10 \mathrm{~mol}\) of argon gas is admitted to an evacuated \(50 \mathrm{~cm}^{3}\) container at \(20^{\circ} \mathrm{C}\). The gas then undergoes an isobaric heating to a temperature of \(300^{\circ} \mathrm{C}\).
a. What is the final volume of the gas?
b. Show the process on a \(p V\) diagram. Include a proper scale on both axes.
Equation Transcription:
Text Transcription:
0.10 mol
50 cm^3
20^circC
300^circC
pV
Questions & Answers
QUESTION:
\(0.10 \mathrm{~mol}\) of argon gas is admitted to an evacuated \(50 \mathrm{~cm}^{3}\) container at \(20^{\circ} \mathrm{C}\). The gas then undergoes an isobaric heating to a temperature of \(300^{\circ} \mathrm{C}\).
a. What is the final volume of the gas?
b. Show the process on a \(p V\) diagram. Include a proper scale on both axes.
Equation Transcription:
Text Transcription:
0.10 mol
50 cm^3
20^circC
300^circC
pV
ANSWER:
Step 1 of 3
We have to find the final pressure of the argon gas in a container.
The initial pressure of \(0.10 \mathrm{~mol}\) of argon gas in a \(50 \mathrm{~cm}^{3}\) container at \(20^{\circ} \mathrm{C}\) can be found using ideal gas law.
\(\begin{gathered}p_{1} V_{1}=n R T_{1} \\p_{1}=\frac{n R T_{1}}{V_{1}}\end{gathered}\)
Where,
\(n\) = 0.10
\(V_{1}=50 \mathrm{~cm}^{3}=50 \times 10^{-6} \mathrm{~m}^{3}\)
\(T_{1}=20^{\circ} \mathrm{C}=20+273=293 \mathrm{~K}\)
\(R=8.31 \mathrm{~J} / \mathrm{mol} \mathrm{K}\)
Thus,
\(p_{1}=\frac{(0.6)(8.31)(293)}{50 \times 10^{-6}}\)
\(=4870 \mathrm{kPa}\)
\(\approx 4900 \mathrm{kPa}\)