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chubhunter [2.5K]
2 years ago
8

What is the boiling point of water when the atmospheric pressure exerted on the water is 81 kPa ?

Chemistry
1 answer:
stellarik [79]2 years ago
5 0

Answer: The boiling point of water at 81 kPa will be 298.17 K.

Explanation:

The boiling point of water is 373 K at atmospheric pressure. So, to calculate the boiling point at 81 kPa, we use the law given by Gay-Lussac.

Gay-Lussac Law states that pressure is directly proportional to the temperature when volume and moles remain constant.

Mathematically,

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

where,

P_1\text{ and }T_1 = Initial pressure and temperature

P_2\text{ and }T_2 = Final pressure and temperature

We are given:

P_1=1atm=101.325kPa\\T_1=373K\\P_2=81kPa\\T_2=?K

Putting values in above equation, we get:

\frac{101.325kPa}{373K}=\frac{81kPa}{T_2}\\\\T_2=298.17K

Hence, the boiling point of water at 81 kPa will be 298.17 K.

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A sample of an ideal gas has a volume of 2.30 L at 281 K and 1.02 atm. 1.76 atm is the pressure when the volume is 1.41 L and the temperature is 298 K.

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\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}

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P_{2} = \frac{1.02\ atm \times 2.30\ L \times 298\ K}{281\ K \times 1.41\ L}

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Learn more about the Combined gas Law here: brainly.com/question/13538773

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