Question

What is the weight of 8.31 liter CH4 in 5 atm and 127°c

Answer 1

Answer:

Convert

Explanation:Convert the right measurement to get the correct answer

Answer 2

The weight of 8.31 liter CH₄ in 5 atm and 127°C is 20 grams

Explanation:

$P=5$ atm = $5\times 10^5$ N/m²

$T=273+127=400$ K

$V=8.31$ L = $8.31\times 10^{-3}$ m³

From gas equation we know that

$PV=nRT$, where R= 8.31 J/mole-K and n is no. of moles

$5\times 10^5\times 8.31\times 10^{-3}=n\times 8.31\times 400$

$n=\frac{5\times 8.31\times 10^2}{8.31\times 400}$

$\implies n=\frac{500}{400}$ moles

$\implies n=1.25$ moles

We know that

No. of moles = Weight / Molecular Weight

Molecular Weight of CH₄ = 16

Therefore,

$1.25=\frac{\text{Weight of }CH_4}{16}$

$\implies \text{Weight of }CH_4=1.25\times 16$

$\implies \text{Weight of }CH_4=20$ gram

Hope this answer is helpful.

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