Respuesta :
Answer:
The pressure of the sample CH₄ gas: P = 0.41 atm
Explanation:
According to the ideal gas equation:
[tex]PV = nRT[/tex]
[tex]or,\; P= \frac{nRT}{V}[/tex]
Here, P is the total pressure of gas
V is the Total Volume of gas
T is Temperature
R is the gas constant = 0.08206 L·atm·K⁻¹·mol⁻¹
n is the number of moles of gas = [tex]\frac{given\: mass\: (w)}{molar\: mass\: (m)}[/tex]
Given: Volume of gas: V = 30 L, Temperature: T = 402 K, Given mass of CH₄ gas: w = 6.022 g, Molar mass of CH₄ gas: m = 16 g/mol
So, number of moles of CH₄: [tex]n = \frac{w}{m} = \frac{6.022\: g}{16\: g/mol} = 0.376\: mol[/tex]
Now, to calculate the total pressure of the sample CH₄ gas, we use the equation,
[tex]P= \frac{nRT}{V}[/tex]
[tex]P = \frac{0.376\: mol \times 0.08206\: L.atm.K^{-1}.mol^{-1} \times 402\: K}{30\: L}[/tex]
[tex]P = 0.41\: atm[/tex]
Therefore, the pressure of the sample CH₄ gas: P = 0.41 atm
The pressure of a sample of CH₄ gas (6.022 g) in a 30L vessel at 402 k is 0.413 atm.
How do we calculate pressure of gas?
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 ideal gas equation, visit the below link:
https://brainly.com/question/24236411
#SPJ3
