Determine the density of methane (CH4) in g/L at 24.7oC and 1.1 atm

ideal gas eqn: P = (rho)RT, where:

pressure: P = 1.1 atm

specific gas constant: R, R = |R / M, where

|R = 0.0821 L-atm / K-mol and,

molecular weightCH4: M = 0.016 kg/mol

R = 0.0821/0.016 = 5.13 and

temp. T = 24.7 + 273 = 291 K, therefore,

density: (rho) = P / RT = 1.1 / (5.13)(291)

(rho) = 62.4 g/L

please tell me if this is correct if not help me to correct my work

No, it isn't correct. First, since you know T to the tenth, I would use 273.2 + 27.4 for T and that gives 297.9 K and not 291 K. Then if I wanted density in g/L I would use the molar mass in grams.

PM = rho*RT and
rho = PM/RT = (1.1*16)/(0.0821*297.9) = 0.7196 g/L which I would round to 0.720 g/L.

Your calculation is mostly correct, but there is a minor mistake in converting the specific gas constant, R.

The specific gas constant is given as 0.0821 L-atm / K-mol, and you correctly divided it by the molecular weight of methane (CH4), which is 0.016 kg/mol. However, when dividing, you accidentally wrote 0.0821/0.016 as 5.13.

To correct it, calculate 0.0821 / 0.016, which gives you 5.13125.

Now, let's recalculate the density of methane using the corrected value of R:

Density (ρ) = Pressure (P) / (R * Temperature (T))

Using the values you provided:
P = 1.1 atm
R = 0.0821 / 0.016 = 5.13125
T = 24.7 + 273 = 297.7 K

ρ = 1.1 / (5.13125 * 297.7)
ρ ≈ 0.004071 g/L

So, the corrected density of methane at 24.7°C and 1.1 atm is approximately 0.004071 g/L.