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Bess [88]
2 years ago
7

A gas that has a volume of 28 Liters, a Temperature of 318 K, and an unknown pressure initially

Chemistry
1 answer:
Kobotan [32]2 years ago
8 0

Answer: 25.07 atm

Explanation:

Given that,

Initial Pressure of gas (P1) = ?

Initial Volume of gas (V1) = 28L

Initial Temperature of gas (T1) = 318K

Final pressure of helium = 20 atm

Final volume of gas (V2) = 34L

Final temperature of gas (T2) = 308K

Since pressure, volume and temperature are given, apply the combined gas equation

(P1V1)/T1 = (P2V2)/T2

(P1 x 28L)/318K = (20 atm x 34L)/308K

28L•P1/318K = 680 atm•L/308K

To get P1, cross multiply

28L•P1 x 308K = 680 atm•L x 318K

8624L•K•P1 = 216240 atm•L•K

Divide both sides by 8624L•K

8624L•K•P1/8624L•K = 216240 atm•L•K/8624L•K

P1 = 25.07 atm

Thus, the original pressure of the gas is 25.07 atmosphere

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This is an incomplete question, here is a complete question.

The Henry's law constant for oxygen dissolved in water is 4.34 × 10⁹ g/L.Pa at 25⁰C.If the partial pressure of oxygen in air is 0.2 atm, under atmospheric conditions, calculate the molar concentration of oxygen in air-saturated and oxygen saturated water.

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Explanation :

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C_{O_2}=(4.34\times 10^9g/L.Pa)\times (1.97\times 10^{-6}Pa)

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x = 23.65 * 10^23

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