150 mL of a gas at S.T.P. were taken to `20^(@)C` and 0.96 bar pressure. What is the change in volume of the gas?

Explanation

To solve the problem of finding the change in volume of a gas when it is taken from STP conditions to a new temperature and pressure, we can use the Ideal Gas Law. Here’s a step-by-step solution:

### Step 1: Understand the conditions at STP
At Standard Temperature and Pressure (STP):
- Standard Temperature (T1) = 0°C = 273 K
- Standard Pressure (P1) = 1 bar

### Step 2: Identify the new conditions
The new conditions given in the problem are:
- New Temperature (T2) = 20°C = 20 + 273 = 293 K
- New Pressure (P2) = 0.96 bar
- Initial Volume (V1) = 150 mL

### Step 3: Use the Ideal Gas Law
The relationship between the initial and final states of the gas can be expressed using the formula:
\[ \frac{P1 \cdot V1}{T1} = \frac{P2 \cdot V2}{T2} \]
Where:
- \( V2 \) is the final volume we need to find.

### Step 4: Rearrange the formula to solve for V2
Rearranging the formula gives us:
\[ V2 = \frac{P1 \cdot V1 \cdot T2}{P2 \cdot T1} \]

### Step 5: Substitute the values into the equation
Now, substituting the known values:
- \( P1 = 1 \, \text{bar} \)
- \( V1 = 150 \, \text{mL} \)
- \( T2 = 293 \, \text{K} \)
- \( P2 = 0.96 \, \text{bar} \)
- \( T1 = 273 \, \text{K} \)

The equation becomes:
\[ V2 = \frac{1 \, \text{bar} \cdot 150 \, \text{mL} \cdot 293 \, \text{K}}{0.96 \, \text{bar} \cdot 273 \, \text{K}} \]

### Step 6: Calculate V2
Calculating the above expression:
\[ V2 = \frac{1 \cdot 150 \cdot 293}{0.96 \cdot 273} \approx \frac{43950}{261.12} \approx 168.2 \, \text{mL} \]

### Step 7: Calculate the change in volume
Now, we find the change in volume:
\[ \text{Change in Volume} = V2 - V1 = 168.2 \, \text{mL} - 150 \, \text{mL} = 18.2 \, \text{mL} \]

### Final Answer
The change in volume of the gas is approximately **18.2 mL**.

---