Question

If a gas sample has a pressure of 30.7 kpa at 0.00, by how many degrees celsius does the temperature have to increase to cause the pressure to double?

Answer 1

Pressure of the gas P1 = 30.7 kpa  
When it doubled P2 = 61.4 kpa  
Temperature T1 = 0 => T1 =. 0 + 273 =273 
Temperature T2 =? 
We have pressure temperature equation P1T1 = P2T2
 => T2 = P1T1 / P2 = (30.7 x 273) / 61.4 = 136.5 
 So the temperature for doubling the pressure is 136.5.

Answer 2

Answer: The temperature has to increase by $273^0C$.

Explanation:

Gay-Lussac's Law: This law states that pressure is directly proportional to the temperature of the gas at constant volume and number of moles.

$P\propto T$     (At constant volume and number of moles)

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

where,

$P_1$ = initial pressure of gas  = 30.7 kPa

$P_2$ = final pressure of gas  = $2\times 30.7=61.4kPa$

$T_1$ = initial temperature of gas  =$0^0C=(0+273)K=273K$

$T_2$ = final temperature of gas = ?

$\frac{30.7}{273}=\frac{61.4}{T_2}$

$T_2=546K=(546-273)^0C=273^0C$

Therefore, the temperature has to increase by $273^0C$  to increase to cause the pressure to double.