One litre of oxygen at a pressure of 1 atm and two litres of nitrogen at a pressure of 0.5 atm are introduced into a vessel of volume 1 litre. If there is no change in temperature, the final pressure of the mixture of gas (in atm) is

Options

A. (a)1.5
B. (b)2.5
C. (c)2
D. (d)4

Answer

To solve the problem, we will use the ideal gas law, which states that:

\[ PV = nRT \]

Where:
- \( P \) = pressure
- \( V \) = volume
- \( n \) = number of moles
- \( R \) = ideal gas constant
- \( T \) = temperature

### Step 1: Determine the number of moles of oxygen (O₂)

Given:
- Volume of oxygen \( V_{O_2} = 1 \) L
- Pressure of oxygen \( P_{O_2} = 1 \) atm

Using the ideal gas law:

\[ n_{O_2} = \frac{P_{O_2} \cdot V_{O_2}}{RT} \]

Substituting the values:

\[ n_{O_2} = \frac{1 \cdot 1}{RT} = \frac{1}{RT} \]

### Step 2: Determine the number of moles of nitrogen (N₂)

Given:
- Volume of nitrogen \( V_{N_2} = 2 \) L
- Pressure of nitrogen \( P_{N_2} = 0.5 \) atm

Using the ideal gas law:

\[ n_{N_2} = \frac{P_{N_2} \cdot V_{N_2}}{RT} \]

Substituting the values:

\[ n_{N_2} = \frac{0.5 \cdot 2}{RT} = \frac{1}{RT} \]

### Step 3: Calculate the total number of moles in the mixture

The total number of moles \( n_{total} \) is the sum of the moles of oxygen and nitrogen:

\[ n_{total} = n_{O_2} + n_{N_2} = \frac{1}{RT} + \frac{1}{RT} = \frac{2}{RT} \]

### Step 4: Determine the final pressure of the gas mixture

The total volume of the vessel is \( V_{mix} = 1 \) L. Using the ideal gas law for the mixture:

\[ P_{mix} \cdot V_{mix} = n_{total} \cdot RT \]

Substituting the values:

\[ P_{mix} \cdot 1 = \left(\frac{2}{RT}\right) \cdot RT \]

This simplifies to:

\[ P_{mix} = 2 \text{ atm} \]

### Final Answer

The final pressure of the mixture of gases is **2 atm**.

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