Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 12

# Find the Loss in the Mass of 1 Mole of an Ideal Monatomic Gas Kept in a Rigid Container as It Cools Down by 100°C. the Gas Constant R = 8.3 J K−1 Mol−1. - Physics

Sum

Find the loss in the mass of 1 mole of an ideal monatomic gas kept in a rigid container as it cools down by 100°C. The gas constant R = 8.3 J K−1 mol−1.

#### Solution

Given: Number of moles of gas, n = 1

Change in temperature, ∆T = 10°C

Energy possessed by a mono atomic gas,

$E = \frac{3}{2}nRdT$

Now,

$R = 8.3 J/mol -K$

This decrease in energy causes loss in mass of the gas. Thus,

Loss in mass, $∆ m = \frac{Q}{c^2}$

$\Rightarrow ∆ m = \frac{1 . 5 \times 8 . 3 \times 10}{c^2}$
$= \frac{124 . 5}{9 \times {10}^{16}} = 1 . 38 \times {10}^{- 15} kg$

Concept: Energy and Momentum
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#### APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 2
Chapter 25 The Special Theory of Relativity
Q 19 | Page 458