# Pressure Temperature Law

Pressure Temperature Law (Amonton’s Law):

According to this Law at constant volume, pressure of a given mass of a gas is directly proportional to absolute Temperature (T).

I.e.

$P \propto T$

$\dfrac{P}{T}= \text{Constant}$

$\dfrac{P_1}{T_1}$ = $\dfrac{P_2}{T_2}$

$P_t = P_0 + \dfrac{P_0}{273} \times P_0 \left( \dfrac{t + 273}{273} \right)$

$P_1 = \dfrac{P_0T}{273}$

Similarly for Charle’s Law

$V_t = V_0 + \dfrac{V_0}{273} \times t = V_0 \left( 1 + \dfrac{t}{273} \right)$

= $V_0 \left( \dfrac{t + 273}{273} \right) = V_0 \dfrac{T}{273}$

Example1:

A balloon blown up has a volume of 500 mL at $5^0 C$. The balloon is distended to 7/8th of its maximum stretching capacity

(i)  Will it burst at $30^0 C$?

(ii) Calculate the minimum temperature above which it will burst?

Solution:

(i) Maximum capacity of the balloon = $500 \times \dfrac{8}{7} = 571.4 mL$

In means the balloon will burst when the volume will be more than 571.4 mL. According to Char1e’s law:

$\dfrac{V_1}{T_1} = \dfrac{V_2}{T_2} \\[3mm] \text{or} \hspace{10mm} \dfrac{500 mL}{273 + 5)K} = \dfrac{V_2}{(273 + 30) K} \\[3mm] \text{or} \hspace{10mm} V_2 = \dfrac{500 mL \times 303 K}{278 K} = 544.96 mL$

Since the volume is less than 571.4 mL hence balloon will not burst at $30^0 C$.

(ii) To calculate minimum temperature above which balloon will burst, again apply Charle’s Law:

$\dfrac{V_1}{T_1} = \dfrac{V_2}{T_2} \\[3mm] \text{or} \hspace{10mm} \dfrac{500 mL}{278 K} = \dfrac{571.4 mL}{T_2} \\[3mm] \text{or} \hspace{10mm} T_2 = \dfrac{571.4 mL \times 278 K}{500 mL} = 317.7 K \\[3mm] = 317.7- 273 = 44.7^0 C$

Hence balloon Will burst at $44.7^0C$.

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