A certain gas occupies 0.418 litres at `27^@C` and 740 mm of Hg.
If we increase the weight of the gas to 7.5 g in the same vessel and the temperature is changed to 280 K, what would be its pressure ?

Explanation

To find the pressure of the gas after increasing its weight to 7.5 g and changing the temperature to 280 K, we can use the Ideal Gas Law, which is expressed as:

\[ PV = nRT \]

Where:
- \( P \) = Pressure of the gas
- \( V \) = Volume of the gas (in liters)
- \( n \) = Number of moles of the gas
- \( R \) = Universal gas constant (0.0821 L·atm/(K·mol))
- \( T \) = Temperature in Kelvin

### Step 1: Calculate the number of moles of the gas
Given that the mass of the gas is 7.5 g and the molar mass of the gas is 181.5 g/mol, we can calculate the number of moles (\( n \)) using the formula:

\[ n = \frac{\text{mass}}{\text{molar mass}} \]

Substituting the values:

\[ n = \frac{7.5 \, \text{g}}{181.5 \, \text{g/mol}} \]

Calculating this gives:

\[ n = 0.0413 \, \text{mol} \]

### Step 2: Identify the volume and temperature
From the question, we know:
- Volume \( V = 0.418 \, \text{L} \)
- Temperature \( T = 280 \, \text{K} \)

### Step 3: Use the Ideal Gas Law to find the pressure
We can rearrange the Ideal Gas Law to solve for pressure \( P \):

\[ P = \frac{nRT}{V} \]

Substituting the known values:

- \( n = 0.0413 \, \text{mol} \)
- \( R = 0.0821 \, \text{L·atm/(K·mol)} \)
- \( T = 280 \, \text{K} \)
- \( V = 0.418 \, \text{L} \)

Now, substituting these values into the equation:

\[ P = \frac{(0.0413 \, \text{mol}) \times (0.0821 \, \text{L·atm/(K·mol)}) \times (280 \, \text{K})}{0.418 \, \text{L}} \]

### Step 4: Calculate the pressure
Calculating the numerator:

\[ 0.0413 \times 0.0821 \times 280 = 0.952 \, \text{L·atm} \]

Now, divide by the volume:

\[ P = \frac{0.952 \, \text{L·atm}}{0.418 \, \text{L}} = 2.28 \, \text{atm} \]

### Final Answer
The pressure of the gas after increasing the weight to 7.5 g and changing the temperature to 280 K is approximately **2.28 atm**.

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