Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11

# A Gas Cylinder Has Walls that Can Bear a Maximum Pressure of 1.0 × 106 Pa. It Contains a Gas at 8.0 × 105 Pa and 300 K. - Physics

Sum

A gas cylinder has walls that can bear a maximum pressure of 1.0 × 106 Pa. It contains a gas at 8.0 × 105 Pa and 300 K. The cylinder is steadily heated. Neglecting any change in the volume, calculate the temperature at which the cylinder will break.

#### Solution

Given:-
Maximum pressure that the cylinder can bear, Pmax = 1.0 × 106 Pa
Pressure in the gas cylinder, P1 = 8.0 × 105 Pa
Temperature in the cylinder, T1 = 300 K

Let T2 be the temperature at which the cylinder will break.
Volume is constant. Thus,         ............(Given)
V1V2 = V

Applying the five variable gas equation, we get

$\frac{P_1 V}{T_1} = \frac{P_2 V}{T_2} .........( \because V_1 = V_2 )$

$\Rightarrow \frac{P_1}{T_1} = \frac{P_2}{T_2}$

$\Rightarrow T_2 = \frac{P_2 \times T_1}{P_1}$

$\Rightarrow T_2 = \frac{1 . 0 \times {10}^6 \times 300}{8 . 0 \times {10}^5}=375K$

Concept: Kinetic Theory of Gases and Radiation - Kinetic Interpretation of Temperature
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#### APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 2
Chapter 2 Kinetic Theory of Gases
Q 7 | Page 34