What is the pressure of a sample of $C H_{4}$ $CH_4$ gas (6.022 g) in a 30.0 L vessel at 402 K?

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What is the pressure of a sample of $C H_{4}$ $CH_4$ gas (6.022 g) in a 30.0 L vessel at 402 K?

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Answer 1 · 2017-02-09T11:09:22

$106.273524 atm$ $"106.273524 atm"$

Explanation:

$Lets recall the gas law$ $" Lets recall the gas law"$

$P V = n R T$ $PV = nRT$

$so P = \frac{(nRT)}{V}$ $"so "P = "(nRT)"/V"$

$Where P = pressure in atm$ $"Where P = pressure in atm"$
$R = the gas constant 0.0821L$ $"R = the gas constant 0.0821L"$
$T = temperature in kelvin$ $"T = temperature in kelvin"$
$n = moles of substance$ $"n = moles of substance "$
$V = volume in litres$ $"V = volume in litres"$

Lets now plug in the variables
R = 0.0821L
T = 402K
now for $n$ $n$ we have to calculate

$n = \frac{amount of substance in grams}{molar mass}$ $"n" = "amount of substance in grams"/"molar mass"$

$n = \frac{6.022g}{16.04 g/mol} = 96.6moles$ $"n" = "6.022g"/ "16.04 g/mol" = "96.6moles"$

V = 30.0L

$\frac{(96.6g * 0.0821 * 402)}{30.0L} = P r e s s u r e$ $"(96.6g * 0.0821 * 402)" / "30.0L" = Pressure$

$\frac{3188.20572}{30.0L} = 106.273524 atm$ $3188.20572 / "30.0L" = "106.273524 atm"$

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What is the pressure of a sample of $C H_{4}$ $CH_4$ gas (6.022 g) in a 30.0 L vessel at 402 K?

1 Answer

Explanation:

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What is the pressure of a sample of CH_4CH4CH_4 gas (6.022 g) in a 30.0 L vessel at 402 K?

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What is the pressure of a sample of $C H_{4}$ $CH_4$ gas (6.022 g) in a 30.0 L vessel at 402 K?