Question

How many grams of oxygen gas are there in a 2.3L tank at 7.5 atm and 24° C?

Answer 1

Answer:

22.656 grams of oxygen gas are there in a 2.3L tank at 7.5 atm and 24° C

Explanation:

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law:

P * V = n * R * T

where R is the molar constant of the gases and n the number of moles.

In this case you know:

• P= 7.5 atm
• V= 2.3 L
• n= ?

• R= 0.082 $\frac{atm*L}{mol*K}$

• T= 24 °C= 297 °K (being 0°C=273°K)

Replacing:

$7.5 atm* 2.3 L=n*0.082 \frac{atm*L}{mol*K} *297K$

Solving:

$n=\frac{7.5 atm* 2.3 L}{0.082 \frac{atm*L}{mol*K} *297K}$

n=0.708 moles

Knowing that oxygen gas is a diatomic gas of molecular form O₂ and its mass is 32 g / mole, you can apply the following rule of three: if 1 mole contains 32 grams, 0.708 moles, how much mass will it have?

$mass=\frac{0.708 moles*32 grams}{1mole}$

mass= 22.656 grams

22.656 grams of oxygen gas are there in a 2.3L tank at 7.5 atm and 24° C

Answer 2

Answer:

The mass of oxygen gas is 22.66 grams

Explanation:

Step 1: Data given

Volume of tank = 2.3 L

Pressure = 7.5 atm

Temperature = 24.0°C = 273 +24 = 297 K

Step 2: Calculate moles oxygen gas

p*V = n*R*T

⇒with p = the pressure of the oxygen gas = 7.5 atm

⇒with V = the volume of the tank : 2.3 L

⇒with n = the moles of oxygen gas = TO BE DETERMINED

⇒with R = the gas constant = 0.08206L*atm/mol*K

⇒with T = the temperature = 297 K

n = (p*V- / (R*T)

n = (7.5 atm * 2.3L) / (0.08206 L*atm/mol*L * 297K)

n = 0.708 moles

Step 3: Calculate mass of oxygen gas

Mass O2 = moles O2 = molar mass O2

Moles O2 = 0.708 moles * 32.0 g/mol

Moles O2 = 22.66 grams

The mass of oxygen gas is 22.66 grams