Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11
Advertisement Remove all ads

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.

Advertisement Remove all ads

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
  Is there an error in this question or solution?
Advertisement Remove all ads

APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 2
Chapter 2 Kinetic Theory of Gases
Q 7 | Page 34
Advertisement Remove all ads
Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×