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(Ill) A sealed container containing 4.0 mol of gas
Chapter 17, Problem 17.48(choose chapter or problem)
(Ill) A sealed container containing 4.0 mol of gas is squeezed, changing its volume from \(0.020 \mathrm{~m}^{3}\) to \(0.018 \mathrm{~m}^{3}\). During this process, the temperature decreases by 9.0 K while the pressure increases by 450 Pa. What was the original pressure and temperature of the gas in the container?
Questions & Answers
QUESTION:
(Ill) A sealed container containing 4.0 mol of gas is squeezed, changing its volume from \(0.020 \mathrm{~m}^{3}\) to \(0.018 \mathrm{~m}^{3}\). During this process, the temperature decreases by 9.0 K while the pressure increases by 450 Pa. What was the original pressure and temperature of the gas in the container?
ANSWER:Step 1 of 4
We are asked to find the original pressure and temperature of the sealed container if it is squeezed to a point where the temperature is decreased by 9K and pressure is increased by 450Pa.
Given data:
\(\begin{array}{l}\text{Initial volume } V_{1}=0.020 \mathrm{~m}^{3}\\
\text{Final Volume } V_{2}=0.018 \mathrm{~m}^{3}\\
\text{Final temperature } =T_{1}-9 \mathrm{~K}\\
\text{Final pressure } =P_{1}+450 \mathrm{~Pa}\\
\text{number of moles, } n=4 \mathrm{~mol}\end{array}\)
Using ideal gas equation:
\(\begin{array}{l}
P V=n R T \\
n=\text { number of moles } \\
P=\text { pressure } \\
V=\text { volume } \\
T=\text { temperature } \\
R=\text { ideal gas } \text { constant }=8.314 \mathrm{~J} / \mathrm{mol}-\mathrm{K}
\end{array}\)