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2 years ago

What is the pressure in a 27.0-L cylinder filled with 44.9 g of oxygen gas at a temperature of 315 KK

Chemistry

1 answer:

steposvetlana [31]2 years ago

6 0

Answer:

1.343 atm

Explanation:

We are given the following;

Pressure, p = ?

Volume v = 27 L

Mass of oxygen = 44.9 g

Temperature, T = 315 K

The formular relating all these variables is the equation;

PV = nRT

where R = gas constant = 0.08206 L atm / mole K

To obtain n, we use;

number of moles, n = Mass / molar mass = 44.9 / 32 = 1.403 moles

From the ideal gas equation;

P = nRT / V

P = 1.403 * 0.08206 * 315 / 27

P = 36.27 / 27 = 1.343 atm

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3 years ago

What is the final pressure (expressed in atm) of a 3.05 l system initially at 724 mm hg and 298 k, that is compressed to a final

krok68 [10]

<u>Answer:</u>

P2 = 778.05 mm Hg = 1.02 atm

<u>Explanation:</u>

We are to find the final pressure (expressed in atm) of a 3.05 liter system initially at 724 mm hg and 298 K which is compressed to a final volume of 2.60 liter at 273 K.

For this, we would use the equation:

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

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Substituting the given values in the equation to get:

$\frac{(724)(3.05)}{298} =\frac{P_2(2.6)}{173}$

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2 years ago