What is the pressure, in mmHg, of a 4.00 g sample of O2 gas, which has a temperature of 37.0 °C, and a volume of 4400 mL?

549.48 mmHg

Explanation:

We'll begin by calculating the number of mole of oxygen in 4 g. This can be obtained as follow:

Molar mass of O₂ = 2 × 16 = 32 g/mol

Mass of O₂ = 4 g

Mole of O₂ =?

Mole = mass /molar mass

Mole of O₂ = 4/32

Mole of O₂ = 0.125 mole

Next, we shall convert 37.0 °C to Kelvin temperature. This can be obtained as follow:

T(K) = T(°C) + 273

T (°C) = 37.0 °C

T(K) = 37.0 °C + 273

T(K) = 310 K

Next, we shall convert 4400 mL to L.

1000 mL = 1 L

Therefore,

4400 mL = 4400 mL × 1 L / 1000 mL

4400 mL = 4.4 L

Next, we shall determine the pressure. This can be obtained as follow:

Number of mole (n) = 0.125 mole

Temperature (T) = 310 K

Volume (V) = 4.4 L

Gas constant (R) = 0.0821 atm.L/Kmol

Pressure (P) =?

PV = nRT

P × 4.4 = 0.125 × 0.0821 × 310

Divide both side by 4.4

P = (0.125 × 0.0821 × 310) / 4.4

P = 0.723 atm

Finally, we shall convert 0.723 atm to mmHg.

1 atm = 760 mmHg

Therefore,

0.723 atm = 0.723 atm × 760 mmHg / 1 atm

0.723 atm = 549.48 mmHg

Thus, the pressure is 549.48 mmHg

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