Question

What is the pressure (in atm) in a 5.00 L tank with 9.00 grams of oxygen gas at 350 K?

Answer 1

Answer:

1.61 atm

Explanation:

Using general gas law equation as follows:

PV = nRT

Where;

P = pressure (atm)

V = volume (L)

n = number of moles (mol)

R = gas law constant (0.0821 Latm/molK)

T = temperature (K)

From the information provided in this question,

V = 5.00L

T = 350K

mass of O2 = 9.00g

Using mole = mass/molar mass

Molar mass of O2 = 16(2) = 32g/mol

mole = 9/32

mole = 0.28mol

Using PV = nRT

P = nRT/V

P = 0.28 × 0.0821 × 350/5

P = 8.0458/5

P = 1.609

Pressure = 1.61 atm

Answer 2

The pressure of the oxygen gas at the given temperature and volume is 1.61 atm.

The given parameters;

• volume of the gas, V = 5 L
• mass of the oxygen gas, m = 9 g
• temperature of the gas, T = 350 K

The number of the moles of the given gas is calculated as follows;

$moles = \frac{m}{M} \\\\moles = \frac{9}{32} \\\\moles = 0.28 \ mol$

The pressure of the gas is calculated as follows;

$PV = nRT\\\\P = \frac{nRT}{V} \\\\P = \frac{0.28 \times 0.0821 \times 350}{5} \\\\P = 1.61 \ atm$

Thus, the pressure of the oxygen gas at the given temperature and volume is 1.61 atm.

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