The density of a gas at `27^(@)C` and `760 mm` pressure is `24`. Calculate the temperature at which it will be `18`, the pressure remaining constant.

Options

A. 6
B. 12
C. 18
D. 24

Answer

To solve the problem, we will use the relationship between density and temperature for a gas at constant pressure. The key idea is that for a given gas at constant pressure, the product of density and temperature remains constant.

### Step-by-Step Solution:

1. **Understand the Given Information:**
- Initial Density (ρ₁) = 24
- Initial Temperature (T₁) = 27°C = 27 + 273 = 300 K
- Final Density (ρ₂) = 18
- Final Temperature (T₂) = ?

2. **Use the Relationship:**
Since pressure is constant, we can use the relationship:
\[ \rho_1 \cdot T_1 = \rho_2 \cdot T_2 \]

3. **Substitute the Known Values:**
Plugging in the values we have:
\[ 24 \cdot 300 = 18 \cdot T_2 \]

4. **Calculate T₂:**
Rearranging the equation to solve for T₂:
\[ T_2 = \frac{24 \cdot 300}{18} \]

5. **Perform the Calculation:**
First, calculate the numerator:
\[ 24 \cdot 300 = 7200 \]
Now divide by 18:
\[ T_2 = \frac{7200}{18} = 400 \text{ K} \]

6. **Convert T₂ to Celsius:**
To convert Kelvin back to Celsius:
\[ T_2 = 400 - 273 = 127°C \]

### Final Answer:
The temperature at which the density will be 18, while keeping the pressure constant, is **127°C**.

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