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. Here’s the step-by-step solution:

### Step 1: Identify the 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 = 0 + 273 = 273 K
- Standard pressure (P2) = 1 bar

### Step 2: Use the Density Formula
The density of a gas can be expressed as:
\[ D = \frac{P \cdot M}{R \cdot T} \]
Where:
- D = Density
- P = Pressure
- M = Molar mass (constant for the same gas)
- R = Ideal gas constant (constant)
- T = Temperature in Kelvin

### Step 3: Set Up the Ratio of Densities
For two different states of the same gas, we can set up the following ratio:
\[ \frac{D1}{D2} = \frac{P1}{P2} \cdot \frac{T2}{T1} \]

### Step 4: Substitute the Known Values
Substituting the known values into the equation:
\[ \frac{5.46}{D2} = \frac{2 \, \text{bar}}{1 \, \text{bar}} \cdot \frac{273 \, \text{K}}{300 \, \text{K}} \]

### Step 5: Simplify the Equation
This simplifies to:
\[ \frac{5.46}{D2} = 2 \cdot \frac{273}{300} \]
Calculating the right side:
\[ \frac{273}{300} = 0.91 \]
Thus:
\[ \frac{5.46}{D2} = 2 \cdot 0.91 = 1.82 \]

### Step 6: Solve for D2
Now, rearranging the equation to solve for D2:
\[ D2 = \frac{5.46}{1.82} \]

### Step 7: Calculate D2
Calculating the value:
\[ D2 \approx 3.00 \, \text{g/dm}^3 \]

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

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