Explanation:
The expression for an Ideal Gas is:
[tex]P.V=n.R.T[/tex] (1)
Where:
[tex]P[/tex] is the pressure of the gas
[tex]V[/tex] is the volume of the gas
[tex]n[/tex] the number of moles of gas
[tex]R[/tex] is the gas constant
[tex]T[/tex] is the absolute temperature of the gas
Finding [tex]V[/tex]:
[tex]V=\frac{n.R.T}{P}[/tex] (2)
If we are told the the amount of gas [tex]n[/tex] and pressure [tex]P[/tex] remain constant, but we increase the temperature [tex]T[/tex] by a factor of three; we will have to rewrite (2) with the new temperature [tex]T_{N}[/tex]:
[tex]T_{N}=3T[/tex]
[tex]V=\frac{n.R.3T}{P}[/tex] (3)
[tex]V=3\frac{n.R.T}{P}[/tex] (4)
Now, if we compare (2) with (4), it is clearly noticeable the volume of the gas has increased by a factor of 3.
The volume will triple ( increase by a factor of three )
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The basic formula of pressure that needs to be recalled is:
Pressure = Force / Cross-sectional Area
or symbolized:
[tex]\large {\boxed {P = F \div A} }[/tex]
P = Pressure (Pa)
F = Force (N)
A = Cross-sectional Area (m²)
Let us now tackle the problem !
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In this problem , we will use Ideal Gas Law as follows:
Given:
Initial Temperature of The Gas = T₁
Final Temperature of The Gas = T₂ = 3T₁
Asked:
Final Volume of The Gas = V₂ = ?
Solution:
[tex]\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}[/tex]
[tex]\frac{PV_1}{T_1} = \frac{PV_2}{3T_1}[/tex]
[tex]\frac{V_1}{1} = \frac{V_2}{3}[/tex]
[tex]3(V_1) = 1(V_2)[/tex]
[tex]V_2 = 3V_1[/tex]
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The volume will triple ( increase by a factor of three )
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Grade: High School
Subject: Physics
Chapter: Pressure
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Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant , Liquid , Pressure, Volume , Ideal , Gas , Law
