Question

How many grams of methane gas (CH4) are in a 15.8-liter sample at 1.2 atmospheres and 27°C? Show all work used to solve this problem

Answer 1

Answer : The amount of methane gas is, 12.316 grams.

Solution : Given,

Using ideal gas equation :

`PV=nRT\\\\PV=(w)/(M)* RT\\\\w=(PVM)/(RT)`

where,

P = pressure of the methane gas= 1.2 atm

V = volume of the methane gas = 15.8 L

T = temperature of the methane gas =
`27^oC=273+27=300K`

n = number of moles of the methane gas

R = gas constant = 0.0821 Latm/moleK

M = molar mass of methane gas = 16 g/mole

w = mass of methane gas

Now put all the given values in the above ideal gas equation, we get

`w=((1.2atm)* (15.8L)* (16g/mole))/((0.0821Latm/moleK )* (300K))=12.316g`

Therefore, the amount of methane gas is, 12.316 grams.

Answer 2

12.31 g

Solution:

Let us suppose that the gas is acting ideally, then according to Ideal Gas Equation,

P V = n R T

Also,

n = Mass / M.mass = m / M

So,

P V = m/M R T

Solving for m,

m = P V M / R T

Data Given:

P = 1.2 atm

V = 15.8 L

M = 16 g/mol

T = 27 C + 273 = 300 K

R = 0.0821 atm.L/mol.K

Putting Values,

m = (1.2 atm * 15.8 L * 16 g/mol) / (0.0821 atm.L/mol.K * 300 K)

m = 12.31 g