The density of a gas at `27^(@)C` and 1 atm is d. At what temperature would its density be 0.75d, if the pressure is kept constant?

Options

A. `20^(@)C`
B. `30^(@)C`
C. 400K
D. 300K

Explanation

To solve the problem, we need to find the temperature at which the density of the gas becomes 0.75d while keeping the pressure constant. We will use the relationship between density and temperature derived from the ideal gas law.

### Step-by-Step Solution:

1. **Understand the Relationship**:
The ideal gas law is given by the equation \( PV = nRT \). We can express density (d) in terms of the molar mass (M) and the ideal gas law. The density of a gas is defined as:
\[ d = \frac{m}{V} \]
where \( m \) is the mass of the gas and \( V \) is its volume.

2. **Express Density in Terms of Temperature**:
Rearranging the ideal gas law, we can relate density to temperature:
\[ d = \frac{PM}{RT} \]
From this equation, we can see that density is inversely proportional to temperature when pressure is constant.

3. **Set Up the Equation**:
Since we know the initial density \( d_1 = d \) at temperature \( T_1 = 27^\circ C \) (which is 300 K) and we want to find the temperature \( T_2 \) at which the density \( d_2 = 0.75d \):
\[ d_1 \cdot T_1 = d_2 \cdot T_2 \]

4. **Substitute Known Values**:
Substitute \( d_1 = d \), \( T_1 = 300 \, K \), and \( d_2 = 0.75d \):
\[ d \cdot 300 = 0.75d \cdot T_2 \]

5. **Cancel Out d**:
Since \( d \) is common in both sides, we can cancel it out:
\[ 300 = 0.75 \cdot T_2 \]

6. **Solve for \( T_2 \)**:
Rearranging gives:
\[ T_2 = \frac{300}{0.75} \]
\[ T_2 = 400 \, K \]

### Conclusion:
The temperature at which the density of the gas would be 0.75d, while keeping the pressure constant, is **400 K**.