Question

What temperature, in C, would be required for 1.70 grams of oxygen gas

(oxygen is diatomic) to occupy 9.75 liters at a pressure of 799 Torr? *

Answer 1

2408.1 °C

Step-by-step explanation:

Given data:

Mass of oxygen = 1.70 g

Volume occupy = 9.75 L

Pressure of gas = 799 torr (799/760 = 1.1 atm)

Temperature = ?

Solution:

Formula:

PV = nRT

First of all we will determine the number of moles of oxygen.

Number of moles = mass/ molar mass

Number of moles = 1.70 g/ 32 g/mol

Number of moles = 0.05 mol

Now we will put the values.

R = general gas constant = 0.0821 atm.L/ mol.K

T = PV/nR

T = 1.1 atm× 9.75 L /0.05 mol ×0.0821 atm.L/ mol.K

T = 10.725 /0.004

T = 2681.25 K

kelvin to °C

2681.25 K - 273.15 = 2408.1 °C