Question

a sample of air is trapped in a cylinder as seen in the image below. the initial volume of the air is 4.11 l with a temperature of 23°c and a pressure of 1.29 atm (image on the left). the volume is decreased to 2.19 l and the air is heated to 150°c (image on the right). what is the pressure in the cylinder on the right (in atm)? atm

Explanation:

Step1: Convert temperatures to Kelvin

The formula to convert Celsius to Kelvin is \(T(K)=T(^{\circ}C)+273.15\).
For initial temperature \(T_1 = 23^{\circ}C\), \(T_1=23 + 273.15=296.15K\).
For final temperature \(T_2 = 150^{\circ}C\), \(T_2=150+273.15 = 423.15K\).

Step2: Use the combined gas law

The combined gas law is \(\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\).
We need to solve for \(P_2\), so \(P_2=\frac{P_1V_1T_2}{V_2T_1}\).
Given \(P_1 = 1.29atm\), \(V_1 = 4.11L\), \(V_2=2.19L\), \(T_1 = 296.15K\), \(T_2 = 423.15K\).
Substitute the values: \(P_2=\frac{1.29\times4.11\times423.15}{2.19\times296.15}\).
First, calculate the numerator: \(1.29\times4.11\times423.15=(1.29\times4.11)\times423.15 = 5.3019\times423.15\approx2243.7\).
Then, calculate the denominator: \(2.19\times296.15 = 648.5685\).
Now, \(P_2=\frac{2243.7}{648.5685}\approx3.46\).

Answer:

\(3.46\)