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` ? .

Explanation

To find the density of the gas at Standard Temperature and Pressure (STP), we can use the relationship between density, pressure, and temperature derived from the Ideal Gas Law. Here’s a step-by-step solution:

### Step 1: Identify Given Values
- Density at given conditions (D1) = 5.46 g/dm³
- Temperature (T1) = 27°C = 300 K (since K = °C + 273)
- Pressure (P1) = 2 bar
- Standard Temperature (T2) = 0°C = 273 K
- Standard Pressure (P2) = 1 bar

### Step 2: Use the Density Formula
The density of a gas can be expressed using the formula:
\[ D = \frac{PM}{RT} \]
Where:
- D = density
- P = pressure
- M = molar mass (constant for the gas)
- R = universal gas constant
- T = temperature in Kelvin

### Step 3: Set Up the Ratio of Densities
We can relate the densities at two different conditions (D1 at P1, T1 and D2 at P2, T2) as follows:
\[ \frac{D2}{D1} = \frac{P2 \cdot T1}{P1 \cdot T2} \]

### Step 4: Substitute the Known Values
Substituting the known values into the equation:
- D1 = 5.46 g/dm³
- P2 = 1 bar
- T1 = 300 K
- P1 = 2 bar
- T2 = 273 K

The equation becomes:
\[ \frac{D2}{5.46} = \frac{1 \cdot 300}{2 \cdot 273} \]

### Step 5: Calculate the Right Side
Calculating the right side:
\[ \frac{D2}{5.46} = \frac{300}{546} \]
\[ \frac{D2}{5.46} = 0.549 \]

### Step 6: Solve for D2
Now, multiply both sides by 5.46 to find D2:
\[ D2 = 5.46 \cdot 0.549 \]
\[ D2 \approx 3.00 \, \text{g/dm}^3 \]

### Final Answer
The density of the gas at STP is approximately **3.00 g/dm³**.

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