Answer:
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
Answer : The amount of methane gas is, 12.316 grams.
Solution : Given,
Using ideal gas equation :
[tex]PV=nRT\\\\PV=\frac{w}{M}\times RT\\\\w=\frac{PVM}{RT}[/tex]
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 = [tex]27^oC=273+27=300K[/tex]
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
[tex]w=\frac{(1.2atm)\times (15.8L)\times (16g/mole)}{(0.0821Latm/moleK )\times (300K)}=12.316g[/tex]
Therefore, the amount of methane gas is, 12.316 grams.
