To solve the problem step by step, we will use the Ideal Gas Law, which states that \( PV = nRT \), where \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the gas constant, and \( T \) is the temperature.
### Step 1: Write down the initial conditions
- Initial pressure, \( P_1 = 1 \, \text{atm} \)
- Initial temperature, \( T_1 = 273 \, \text{K} \)
- Let the initial number of moles of nitrogen be \( n_1 \).
### Step 2: Write down the final conditions
- Final pressure, \( P_2 = 0.5 \, \text{atm} \)
- Final temperature, \( T_2 = 546 \, \text{K} \)
- Let the final number of moles of nitrogen be \( n_2 \).
### Step 3: Set up the Ideal Gas Law equations
For the initial state:
\[
P_1 V = n_1 R T_1 \quad \text{(1)}
\]
For the final state:
\[
P_2 V = n_2 R T_2 \quad \text{(2)}
\]
### Step 4: Divide the two equations
Dividing equation (1) by equation (2):
\[
\frac{P_1 V}{P_2 V} = \frac{n_1 R T_1}{n_2 R T_2}
\]
This simplifies to:
\[
\frac{P_1}{P_2} = \frac{n_1 T_1}{n_2 T_2}
\]
### Step 5: Substitute the known values
Substituting the known values into the equation:
\[
\frac{1}{0.5} = \frac{n_1 \cdot 273}{n_2 \cdot 546}
\]
This simplifies to:
\[
2 = \frac{n_1 \cdot 273}{n_2 \cdot 546}
\]
### Step 6: Rearranging the equation
Rearranging gives:
\[
n_2 = \frac{n_1 \cdot 273}{2 \cdot 546}
\]
Simplifying further:
\[
n_2 = \frac{n_1}{4}
\]
### Step 7: Calculate the percentage of nitrogen remaining
If we consider the initial amount of nitrogen \( n_1 \) to be 100%, then:
\[
n_2 = \frac{1}{4} n_1 = \frac{1}{4} \cdot 100\% = 25\%
\]
Thus, the percentage of nitrogen remaining in the beaker is \( 25\% \).
### Step 8: Identify \( m \) and \( n \)
The percentage of nitrogen remaining is \( 25\% \), which can be expressed as \( mn\% \) where \( m = 2 \) and \( n = 5 \).
### Step 9: Calculate \( m + n \)
Thus, \( m + n = 2 + 5 = 7 \).
### Final Answer
The value of \( m + n \) is \( 7 \).
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