A gas is filled in a container at certain temperature and pressure. At the same temperature more gas is filled in the vessel. Calculate the percentage increase in the mass of the gas. If the ratio of initial and final pressure is `1:2`.

Answer

To solve the problem, we need to find the percentage increase in the mass of the gas when the pressure of the gas in a fixed volume container is doubled while maintaining the same temperature. Here’s how we can approach the solution step by step.

### Step 1: Understand the relationship between pressure and mass
According to the ideal gas law, the pressure (P) of a gas is directly proportional to the number of moles (n) of gas when the volume (V) and temperature (T) are constant. This can be expressed as:
\[ P \propto n \]
Since the mass (m) of the gas is related to the number of moles by the molar mass (M), we can also say:
\[ P \propto m \]
Thus, if the pressure doubles, the mass of the gas must also double.

### Step 2: Set up the initial and final conditions
Let:
- \( m_i \) = initial mass of the gas
- \( m_f \) = final mass of the gas
- Given that the ratio of initial pressure to final pressure is \( 1:2 \):
\[ \frac{P_i}{P_f} = \frac{1}{2} \]

### Step 3: Relate initial and final mass
From the relationship established in Step 1, we can express the relationship between the initial and final mass:
\[ \frac{m_i}{m_f} = \frac{P_i}{P_f} = \frac{1}{2} \]
This implies:
\[ m_f = 2 m_i \]

### Step 4: Calculate the percentage increase in mass
The percentage increase in mass can be calculated using the formula:
\[ \text{Percentage Increase} = \frac{m_f - m_i}{m_i} \times 100\% \]

Substituting \( m_f = 2 m_i \):
\[ \text{Percentage Increase} = \frac{(2 m_i - m_i)}{m_i} \times 100\% \]
\[ = \frac{m_i}{m_i} \times 100\% \]
\[ = 100\% \]

### Final Answer
The percentage increase in the mass of the gas is **100%**.

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