The Density of an Ideal Gas is 1.25 × 10−3 G Cm−3 at Stp. Calculate the Molecular Weight of the Gas. - Physics

Sum

The density of an ideal gas is 1.25 × 10−3 g cm−3 at STP. Calculate the molecular weight of the gas.

Use R=8.31J K-1 mol-1

Solution

Let:
m = Mass of the gas
M = Molecular mass of the gas
Now,
Density of ideal gas,$\rho$= 1.25 × 10−3 gcm−3 =1.25 kgm−3
Pressure, = 1.01325$\times$105 Pa   (At STP)
Temperature, T = 273 K    (At STP)
Using the ideal gas equation, we get

$PV = nRT . . . (1)$

$n = \frac{m}{M} . . . (2)$

$\therefore PV = \frac{m}{M}RT$

$\Rightarrow M = \frac{m}{V}\frac{RT}{P}$

$\Rightarrow M = \rho\frac{RT}{P}$

$\Rightarrow M = 1 . 25 \times \frac{8 . 31 \times 273}{{10}^5}$

$\Rightarrow M = 2 . 83 \times {10}^{- 2}$

$= 28 . 3 g - {\text { mol }}^{- 1}$

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APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 2
Chapter 2 Kinetic Theory of Gases
Q 9 | Page 34