Answer: 0.42 atm
Explanation:
For this problem, we are going to use the ideal gas law for pressure, which is [tex]P=\frac{nRT}{V}[/tex]. If you do not know yet, R is the ideal gas constant, [tex]0.0821\frac{L*atm}{mol*K}[/tex]First, we must find the number of CH₄ moles.
[tex]6.022gCH_{4} *\frac{1molCH_{4} }{16.043gCH_{4} } =0.38molCH_{4}[/tex]
Next, we will plug in our given information to the ideal gas law for pressure.
[tex]P=\frac{nRT}{V}[/tex]
[tex]P=\frac{0.38mol*0.0821\frac{L*atm}{mol*K} *402K}{30.0L} =0.42 atm[/tex]
The answer to this, according to the work done above, is 0.42 atm.
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The pressure of a sample of CH4 gas (6.022 g) in a 30.0 L vessel at 402 K is 0.413 atm.
Pressure of gas will be calculated by using the ideal gas equation as:
PV = nRT, where
P = pressure of gas = ?
V = volume of gas = 30L
R = universal gas constant = 0.082 L·atm·K⁻¹·mol⁻¹
T = temperature of gas = 402 K
n is moles of gas & it will be calculated as:
n = W/M, where
W = given mass of CH₄ = 6.022g
M = molar mass of CH₄ = 16g/mol
n = 6.022 / 16 = 0.376mol
Now putting all these values on the above equation, we get
P = (0.376)(0.082)(402) / (30)
P = 0.413 atm
Hence required pressure of methane gas is 0.413 atm.
To know more about the ideal gas equation, visit the below link:
brainly.com/question/24236411
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