Question

Let 4 moles of methanol (liquid) combust in 3 moles of gaseous oxygen to form gaseous carbon dioxide and water vapor. Suppose this occurs in a chamber of fixed volume and fixed temperature. If the original pressure is 1.0 atm, what is the final pressure in the chamber. Express your answer in atm.

Answer 1

Answer:

The final pressure is 2.0 atm

Explanation:

Step 1: Data given

Number of moles of methanol = 4 moles

Number of moles of oxygen = 3 moles

Volume and temperature are fixed

Original pressure = 1.0 atm

Step 2: The balanced equation

2 CH3OH + 3 O2 → 2 CO2 + 4 H2O  

For 2 moles of CH3OH consumed, we need 3 moles of O2 to produce 2 moles of CO2 and 4 moles of H2O

This means the mol ratio methanol to oxygen = 2:3

Step 3 : Calculate the fina lpressure

Since methanol is a liquid and not a gas, it doesn't count for the pressure.

Number of moles O2 = 3 mol

This means the total number of moles for the reactants = 3 moles

The total number of moles  for the products = 2 + 4 = 6 moles of gas

Since V and T are fixed (and R is constant)

⇒ we can write the gas the gas law as following:

P2/P1 = n2/n1   OR P2 = (n2*P1)/n1

P2 = (6 mol * 1.0 atm ) / 3.0 mol

P2 = 2.0 atm

The final pressure is 2.0 atm

Answer 2

The final pressure in the chamber is 2.0 atm

Calculating pressure

From the question, we are to determine the final pressure in the chamber

Since, the volume and temperature are fixed,

Then, we can write that

$\frac{P_{1} }{n_{1} } =\frac{P_{2} }{n_{2} }$

Where $P_{1}$ is the initial pressure

$n_{1}$ is the initial number of moles of gases in the chamber

$P_{2}$ is the final pressure

and $n_{2}$ is the final number of moles of gases in the chamber

Now, we will determine the initial and final number of moles of gases present in the chamber

The equation of the reaction is

$2CH_{3} OH(l) + 3O_{2} (g) \rightarrow 2CO_{2} (g) + 4H_{2} O(g)$

This means, 2 moles of methanol will react with 3 moles of oxygen to produce 2 moles of carbon dioxide and 4 moles of water vapour

Therefore, 2 moles, out of the 4 moles of methanol present, will react completely with the 3 moles of gaseous oxygen to give 2 moles of carbon dioxide and 4 moles of water vapour

Before the reaction

Number of moles of gases present = 3 moles

NOTE: Methanol is a liquid

After the reaction

Number of moles of gases present = 2 + 4 = 6 moles

∴ n₁ = 3 moles

n₂ = 6 moles

From the given information,

$P_{1}$ = 1.0 atm

Now, put the parameters into the formula,
$\frac{P_{1} }{n_{1} } =\frac{P_{2} }{n_{2} }$

$\frac{1}{3} = \frac{P_{2} }{6}$

$3P_{2}=6$

$P_{2}= \frac{6}{3}$

$P_{2} = 2.0 \ atm$

Hence, the final pressure in the chamber is 2.0 atm

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