An ideal gas with a fixed number of molecules is maintained at a constant pressure. At `30.0^@ C`, the volume of the gas is `1.50 m^3` . What is the volume of the gas when the temperature is increased to `75.0^@`C?

Options

A. `0.60 m^3`
B. `3.75 m^3`
C. `1.72 m^3`
D. `1.30 m^3`

Answer

To solve the problem, we will use the relationship between the volume and temperature of an ideal gas at constant pressure, which can be derived from the ideal gas law.

### Step-by-Step Solution:

1. **Understand the Given Information**:
- Initial temperature (T1) = 30.0°C
- Initial volume (V1) = 1.50 m³
- Final temperature (T2) = 75.0°C
- Pressure is constant.

2. **Convert Temperatures to Kelvin**:
- The temperature in Kelvin is given by the formula:
\[ T(K) = T(°C) + 273.15 \]
- For T1:
\[ T1 = 30.0 + 273.15 = 303.15 \, K \]
- For T2:
\[ T2 = 75.0 + 273.15 = 348.15 \, K \]

3. **Use the Volume-Temperature Relationship**:
- At constant pressure, the ratio of volume to temperature is constant:
\[ \frac{V1}{T1} = \frac{V2}{T2} \]
- Rearranging gives:
\[ V2 = V1 \times \frac{T2}{T1} \]

4. **Substitute the Known Values**:
- Substitute V1, T1, and T2 into the equation:
\[ V2 = 1.50 \, m^3 \times \frac{348.15 \, K}{303.15 \, K} \]

5. **Calculate V2**:
- First, calculate the fraction:
\[ \frac{348.15}{303.15} \approx 1.149 \]
- Now calculate V2:
\[ V2 \approx 1.50 \, m^3 \times 1.149 \approx 1.724 \, m^3 \]

6. **Final Result**:
- Therefore, the volume of the gas when the temperature is increased to 75.0°C is approximately:
\[ V2 \approx 1.72 \, m^3 \]