A cylinder contains a gas at temperature of $27^{0} C$ and a pressure 1MPa. If the temperature of the gas is lowered to $23^{0} C$ , the change in pressure is :

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Solution

Given,

Intimal pressure, $P_{1} = 1 M p a$

Initial and final temperature, $T_{1} = 27 + 273 = 300 K a n d T_{2} = 23 + 273 = 296 K$

From ideal gas equation

$P α T$ (At constant volume)

$P_{2} = \frac{T_{2}}{T_{1}} P_{1} = \frac{296}{300} \times 1 = 0.98 M p a$

Change in pressure, $P_{2} - P_{1} = 1 - 0.986 = 0.014 M p a$

Hence, change in pressure is $0.014 M p a$

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Q. A cylinder contains a gas at temperature of

27^{o} C

and a pressure

I M P a

. If the temperature of the gas Is lowered to

- 23^{o} C

, the change in pressure is:

A gas cylinder has walls that can bear a maximum pressure of $1.0 \times 10^{6}$ Pa. It contains a gas at $8.0 \times 10^{5}$ Pa and 300 K. The cylinder is steadily heated. Neglecting any change in the volume, calculate the temperature at which the cylinder will break.