Question

To identify a diatomic gas (X2), a researcher carried out the following experiment: She weighed an empty 4.1-L bulb, then filled it with the gas at 2.00 atm and 24.0 ∘C and weighed it again. The difference in mass was 9.5 g . Identify the gas.

Answer 1

Answer:

Nitrogen, $N_2$

Explanation:

Hello,

This is a clear example of what the ideal gas equation is used for, thus, from its mathematical definition:

$PV=nRT$

One can spell it out in terms of mass and molar mass:

$PV=\frac{m}{M}RT$

Now, solving for the molecular mass, $M$:

$M=\frac{mRT}{PV} =\frac{9.5g*0.082\frac{atm*L}{mol*K}*297.15K}{2.00atm*4.1L}\\ M=28.23g/mol$

Now, by taking into account that the gas is diatomic, the matching gas turns out to be nitrogen.

Best regards.

Answer 2

Answer:

The gas is nitrogen gas N2

Explanation:

Step 1: Data given

Volume of the gas = 4.1L

Pressure = 2.00 atm

Temperature = 24.0 °C

Mass = 9.5 grams

Step 2: Calculate moles of gas

p*V = n*R*T

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

⇒ with p = the pressure of the diatomic gas = 2.00 atm

⇒ with V = the volume of the diatomic gas = 4.1L

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

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

⇒ with T = the temperature = 24.0 °C = 297 K

n = (2.00*4.1)/(0.08206*297)

n = 0.336 moles

Step 3: Calculate molar mass

Molar mass = mass / moles

Molar mass = 9.5 grams / 0.336 moles

Molar mass = 28 g/mol = X2

X has a molar mass of 14 g/mol

The gas is nitrogen gas N2