A definite quatity of an ideal gas has a volume of 30 liter at `20^(@)C` The gas is first compressed at `20^(@)C` until the pressure has doubled and then the temperature is raised to `100^(@)C` keeping pressure constant Determine the final volume of gas .

Answer

To solve the problem step by step, we will use the Ideal Gas Law and the principles of gas behavior under changing conditions.

### Step 1: Understand the Initial Conditions
We have a definite quantity of an ideal gas with:
- Initial Volume (V1) = 30 liters
- Initial Temperature (T1) = 20°C = 20 + 273 = 293 K (convert to Kelvin)

### Step 2: Determine the Effect of Doubling the Pressure
The gas is compressed at constant temperature (20°C) until the pressure doubles. According to Boyle's Law, for a given amount of gas at constant temperature:
\[ P_1 V_1 = P_2 V_2 \]
If the initial pressure is \( P \), after doubling, the final pressure \( P_2 = 2P \).

Using the equation:
\[ P \cdot 30 = 2P \cdot V_2 \]
We can cancel \( P \) from both sides (assuming \( P \neq 0 \)):
\[ 30 = 2V_2 \]
\[ V_2 = \frac{30}{2} = 15 \text{ liters} \]

### Step 3: Determine the Final Volume at Increased Temperature
Now, the temperature is raised to 100°C while keeping the pressure constant. Convert the temperature to Kelvin:
- Final Temperature (T2) = 100°C = 100 + 273 = 373 K

Now, we can use Charles's Law, which states that at constant pressure, the volume of a gas is directly proportional to its temperature:
\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]

Where:
- \( V_1 \) = 15 liters (from Step 2)
- \( T_1 \) = 293 K
- \( T_2 \) = 373 K

Substituting the known values:
\[ \frac{15}{293} = \frac{V_f}{373} \]

### Step 4: Solve for Final Volume (V_f)
Cross-multiplying gives:
\[ 15 \cdot 373 = V_f \cdot 293 \]
\[ V_f = \frac{15 \cdot 373}{293} \]

Calculating this:
\[ V_f = \frac{5595}{293} \approx 19.1 \text{ liters} \]

### Final Answer
The final volume of the gas after the changes is approximately **19.1 liters**.

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