Question

The pressure of a sample of ch4 gas (6.022 g) in a 30.0 l vessel at 402 k is __________ atm

Answer 1

Answer:

The pressure of the sample CH₄ gas: P = 0.41 atm

Explanation:

According to the ideal gas equation:

$PV = nRT$

$or,\; P= \frac{nRT}{V}$

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 = $\frac{given\: mass\: (w)}{molar\: mass\: (m)}$

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: $n = \frac{w}{m} = \frac{6.022\: g}{16\: g/mol} = 0.376\: mol$

Now, to calculate the total pressure of the sample CH₄ gas, we use the equation,

$P= \frac{nRT}{V}$

$P = \frac{0.376\: mol \times 0.08206\: L.atm.K^{-1}.mol^{-1} \times 402\: K}{30\: L}$

$P = 0.41\: atm$

Therefore, the pressure of the sample CH₄ gas: P = 0.41 atm

Answer 2

From  ideal  gas  equation,  that  is   PV =nRT
P=nRT/V
n  is  the  number  of  moles
the molar  mass  of  CH4 =12+4=16
R  is the gas  contant (0.0821 atm/mol/k)
n= 6.022/16=0.376 moles
P={(0.376 moles  x 0.0821 atm/mol/k  x 402k) /30.0L} =0.414 atm