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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