A container having 3 mole of ideal gas occupies 60 litres at pressure P and temperature T. If 0.1 mole of gas is introduced at same P and T in container the change in volume will be ________ litre.

Explanation

To solve the problem step by step, we will use the ideal gas law and the relationship between volume and the number of moles of gas.

### Step 1: Understand the Ideal Gas Law
The ideal gas law is given by the equation:
\[ PV = nRT \]
Where:
- \( P \) = pressure
- \( V \) = volume
- \( n \) = number of moles
- \( R \) = gas constant
- \( T \) = temperature

In this case, both pressure \( P \) and temperature \( T \) are constant.

### Step 2: Establish the Relationship Between Volume and Moles
Since \( P \) and \( T \) are constant, we can say that the volume \( V \) is directly proportional to the number of moles \( n \):
\[ V \propto n \]
This implies:
\[ \frac{V_1}{V_2} = \frac{N_1}{N_2} \]
Where:
- \( V_1 \) = initial volume
- \( V_2 \) = final volume
- \( N_1 \) = initial number of moles
- \( N_2 \) = final number of moles

### Step 3: Identify Initial and Final Conditions
From the problem:
- Initial moles \( N_1 = 3 \) moles
- Initial volume \( V_1 = 60 \) liters
- Additional moles introduced = 0.1 moles
- Final moles \( N_2 = N_1 + 0.1 = 3 + 0.1 = 3.1 \) moles

### Step 4: Calculate the Final Volume \( V_2 \)
Using the relationship established in Step 2:
\[ \frac{V_1}{V_2} = \frac{N_1}{N_2} \]
Substituting the known values:
\[ \frac{60}{V_2} = \frac{3}{3.1} \]

Cross-multiplying to solve for \( V_2 \):
\[ 60 \times 3.1 = 3 \times V_2 \]
\[ 186 = 3V_2 \]
\[ V_2 = \frac{186}{3} = 62 \text{ liters} \]

### Step 5: Calculate the Change in Volume \( \Delta V \)
The change in volume is given by:
\[ \Delta V = V_2 - V_1 \]
Substituting the values we found:
\[ \Delta V = 62 - 60 = 2 \text{ liters} \]

### Final Answer
The change in volume will be **2 liters**.

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