Browse Questions

# 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

$PV=nRT$, and $\large\frac{m}{M} =n$, where$m=$ mass in grams, and $M=$ molar mass, we get, $\large\frac{ mRT^2}{P} = \frac{M^2PV^2}{mR}$

$PV=nRT$, 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.