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3 years ago

5

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 t

he pressure to double?

2 answers:

Otrada [13]3 years ago

8 0

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.

faust18 [17]3 years ago

6 0

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.

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