What is the pressure in a 15.0-L cylinder filled with 32.7 g of oxygen gas at a temperature of 302 K

1.68 atm

Explanation:

First we must obtain the number of moles of the oxygen = 32.7g/32 g/mol = 1.02 moles of oxygen

V= 15.0 L

T= 302 K

R = 0.082 atmLmol-1K-1

From the ideal gas equation;

PV=nRT

P= nRT/V

P= 1.02 *  0.082 * 302/15.0

P= 1.68 atm

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