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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Kinetic Theory of Gases
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The quantity $\large\frac {mRT^2}{\rho}$ of an ideal gas is proportional to the:

(A) square of pressure (B) square of volume (C) inverse of pressure (D) inverse of volume
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2 Answers

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$PV=nRT$, and $\large\frac{m}{M} =n$, where$ m=$ mass in grams, and $M=$ molar mass, we get, $\large\frac{ mRT^2}{P} = \frac{M^2PV^2}{mR}$
answered Jan 22, 2014 by balaji.thirumalai
 
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$PV=nRT$, and $\large\frac{m}{M} =n$, where$ m=$ mass in grams, and $M=$ molar mass, we get, $\large\frac{ mRT^2}{P} = \frac{M^2PV^2}{mR}$
Hence its proportional to the square of the voluume.
answered Jan 22, 2014 by balaji.thirumalai
 

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