Density of a gas is found to be `5.46//dm^(3)` at `27^(@)C` at 2 bar pressure What will be its density at `STP` ? .

Options

A. ` 3.0 g dm^(-3)`
B. `5.0 g dm^(-3)`
C. `6.0 g dm^(-3)`
D. `10.82 g dm^(-3)`

Explanation

To find the density of a gas at Standard Temperature and Pressure (STP) given its density at a different temperature and pressure, we can use the relationship derived from the ideal gas law.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Density at initial conditions (d1) = 5.46 g/dm³
- Initial temperature (T1) = 27°C = 27 + 273 = 300 K
- Initial pressure (P1) = 2 bar
- Standard temperature (T2) = 0°C = 273 K
- Standard pressure (P2) = 1 bar

2. **Use the Ideal Gas Law in Density Form:**
The ideal gas equation can be rearranged to express density:
\[ P = dRT/M \]
where \( P \) is pressure, \( d \) is density, \( R \) is the gas constant, \( T \) is temperature, and \( M \) is molar mass. For two different states, we can set up the following equations:
\[ P_1 = d_1 \cdot R \cdot T_1 / M \quad \text{(1)} \]
\[ P_2 = d_2 \cdot R \cdot T_2 / M \quad \text{(2)} \]

3. **Divide the Two Equations:**
Dividing equation (1) by equation (2) eliminates \( R \) and \( M \):
\[ \frac{P_1}{P_2} = \frac{d_1 \cdot T_1}{d_2 \cdot T_2} \]

4. **Rearrange to Solve for \( d_2 \):**
Rearranging gives:
\[ d_2 = d_1 \cdot \frac{P_2 \cdot T_1}{P_1 \cdot T_2} \]

5. **Substitute the Known Values:**
Substitute \( d_1 = 5.46 \, \text{g/dm}^3 \), \( P_1 = 2 \, \text{bar} \), \( P_2 = 1 \, \text{bar} \), \( T_1 = 300 \, \text{K} \), and \( T_2 = 273 \, \text{K} \):
\[ d_2 = 5.46 \cdot \frac{1 \cdot 300}{2 \cdot 273} \]

6. **Calculate \( d_2 \):**
\[ d_2 = 5.46 \cdot \frac{300}{546} = 5.46 \cdot \frac{1}{2} = 2.73 \, \text{g/dm}^3 \]

7. **Final Calculation:**
\[ d_2 = 3 \, \text{g/dm}^3 \]

### Conclusion:
The density of the gas at STP is **3 g/dm³**.

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