Nitrogen gas is kept in an open beaker at 273 K and 1 atm pressure. If the pressure of the surrounding suddenly falls to 0.5 atm and the temperature increases to 546 K, then the percentage of nitrogen remaining in the beaker is `mn%` of the initial amount. Then the value of m+n is:

Answer

To solve the problem, we will use the ideal gas law and the concept of moles of gas under different conditions.

### Step-by-Step Solution:

1. **Identify Initial Conditions**:
- Initial Temperature (T1) = 273 K
- Initial Pressure (P1) = 1 atm
- Initial number of moles of nitrogen (N1) = N (assume N moles)

2. **Identify Final Conditions**:
- Final Temperature (T2) = 546 K
- Final Pressure (P2) = 0.5 atm
- Final number of moles of nitrogen (N2) = ?

3. **Use the Ideal Gas Law**:
The ideal gas law states that \( PV = nRT \). Since the volume (V) is constant (the beaker is open), we can set up the equation as follows:
\[ \frac{P_1 V}{N_1 T_1} = \frac{P_2 V}{N_2 T_2} \]
Since V is constant, we can cancel it out:
\[ \frac{P_1}{N_1 T_1} = \frac{P_2}{N_2 T_2} \]

4. **Rearranging the Equation**:
Rearranging gives us:
\[ N_2 = N_1 \cdot \frac{P_2 T_1}{P_1 T_2} \]

5. **Substituting Values**:
Substitute the known values into the equation:
\[ N_2 = N \cdot \frac{0.5 \cdot 273}{1 \cdot 546} \]
Simplifying this:
\[ N_2 = N \cdot \frac{0.5 \cdot 273}{546} \]
\[ N_2 = N \cdot \frac{136.5}{546} \]
\[ N_2 = N \cdot \frac{1}{4} = 0.25N \]

6. **Calculating the Percentage of Nitrogen Remaining**:
The percentage of nitrogen remaining in the beaker is given by:
\[ \text{Percentage} = \left(\frac{N_2}{N_1}\right) \times 100 = \left(\frac{0.25N}{N}\right) \times 100 = 25\% \]

7. **Identifying m and n**:
From the percentage of nitrogen remaining, we have:
- \( mn\% = 25\% \)
- Here, \( m = 2 \) and \( n = 5 \).

8. **Calculating m + n**:
\[ m + n = 2 + 5 = 7 \]

### Final Answer:
The value of \( m + n \) is **7**.