Solution Found!
Neon Gas Pressure Change in 0.75-L Cylinder
Chapter 6, Problem 6.28(choose chapter or problem)
A 2.50-L sample of neon at \(0.00^{\circ} \mathrm{C}\) and 1.00 atm is compressed into a 0.75-L cylinder. What pressure will the gas exert in the cylinder at \(30.^{\circ} \mathrm{C}\)?
Questions & Answers
QUESTION:
A 2.50-L sample of neon at \(0.00^{\circ} \mathrm{C}\) and 1.00 atm is compressed into a 0.75-L cylinder. What pressure will the gas exert in the cylinder at \(30.^{\circ} \mathrm{C}\)?
ANSWER:Step 1 of 4
To calculate the pressure that the gas will exert in the \(0.75\,L\) cylinder at \({30^ \circ }C\) , we can use the combined gas law equation, which incorporates changes in both temperature and volume while keeping the amount of gas constant.
The combined gas law’s equation is:
\(\dfrac{{{P_1}{V_1}}}{{{T_1}}} = \dfrac{{{P_2}{V_2}}}{{{T_2}}}\)
Where:
\(Initial\,volume\,{V_1} = 2.5\,L\)
\(Initial\,pressure\,{P_1} = 1.00\,atm\)
\(Initial\,temperature\,in\,Kelvin\,{T_1} = 273.15\,K\left( {{{0.00}^ \circ }C{\rm{ + 273}}{\rm{.15}}} \right)\)
\(Final\,volume\,{V_2} = 0.75\,L\)
\(Final\,pressure\,{P_2} = Unknown\)
\(Final\,temperature\,in\,Kelvin\,{T_2} = 303.15\,K\left( {{{30}^ \circ }C{\rm{ + 273}}{\rm{.15}}} \right)\)
Watch The Answer!
Neon Gas Pressure Change in 0.75-L Cylinder
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In this problem, we are tasked with determining the pressure of neon gas in a 0.75-L cylinder at 30°C after compressing it from a 2.50-L sample at 0.00°C and 1.00 atm using the combined gas law equation, with step-by-step calculations leading to a final pressure value of approximately 3.699 atm.