A cylinder containing 20kg of compressed nitrogen at pressure 9 times that of the atmosphere is kept in a store at `7^(@)C`. The safety valve allows nitrogen to escape when it pressure exceeds 10 times that of the atmosphere. How much nitrogen will escape if the temperature rises to `47^(@)C` ?

Answer

To solve the problem step by step, we will use the ideal gas law and the information provided in the question.

### Step 1: Understand the Initial Conditions
We have a cylinder containing 20 kg of nitrogen gas at a pressure of 9 times the atmospheric pressure (P_initial = 9P₀) and a temperature of 7°C.

### Step 2: Convert Temperature to Kelvin
To use the ideal gas law, we need to convert the temperature from Celsius to Kelvin.
- Initial temperature (T_initial) = 7°C = 7 + 273 = 280 K
- Final temperature (T_final) = 47°C = 47 + 273 = 320 K

### Step 3: Calculate the Number of Moles of Nitrogen
The molecular mass of nitrogen (N₂) is 28 g/mol.
- Convert the mass of nitrogen to grams:
- 20 kg = 20,000 g
- Calculate the number of moles (n_initial):
\[ n_{\text{initial}} = \frac{\text{mass}}{\text{molecular mass}} = \frac{20000 \, \text{g}}{28 \, \text{g/mol}} \approx 714.29 \, \text{mol} \]

### Step 4: Apply the Ideal Gas Law for Initial Conditions
Using the ideal gas law \( PV = nRT \):
- For the initial state:
\[ P_{\text{initial}} V = n_{\text{initial}} R T_{\text{initial}} \]
\[ 9P₀ V = 714.29 R \cdot 280 \]

### Step 5: Apply the Ideal Gas Law for Final Conditions
For the final state, the maximum pressure allowed is 10 times the atmospheric pressure (P_final = 10P₀):
\[ P_{\text{final}} V = n_{\text{final}} R T_{\text{final}} \]
\[ 10P₀ V = n_{\text{final}} R \cdot 320 \]

### Step 6: Relate Initial and Final Conditions
Since the volume (V) and the gas constant (R) are the same for both states, we can equate the two equations:
\[ 9P₀ V = 714.29 R \cdot 280 \]
\[ 10P₀ V = n_{\text{final}} R \cdot 320 \]

### Step 7: Solve for Final Moles of Nitrogen
From the first equation, we can express \( V \) in terms of \( P₀ \):
\[ V = \frac{714.29 R \cdot 280}{9P₀} \]
Substituting this into the second equation:
\[ 10P₀ \left(\frac{714.29 R \cdot 280}{9P₀}\right) = n_{\text{final}} R \cdot 320 \]
This simplifies to:
\[ \frac{10 \cdot 714.29 \cdot 280}{9} = n_{\text{final}} \cdot 320 \]
Solving for \( n_{\text{final}} \):
\[ n_{\text{final}} = \frac{10 \cdot 714.29 \cdot 280}{9 \cdot 320} \]

### Step 8: Calculate the Final Mass of Nitrogen
Now, we can find the final mass of nitrogen:
\[ \text{mass}_{\text{final}} = n_{\text{final}} \cdot 28 \]

### Step 9: Calculate the Mass of Nitrogen Escaped
The mass of nitrogen that escapes is:
\[ \text{mass}_{\text{escaped}} = \text{mass}_{\text{initial}} - \text{mass}_{\text{final}} = 20 \, \text{kg} - \text{mass}_{\text{final}} \]

### Final Calculation
After performing the calculations through the steps above, we find that approximately 0.6 kg of nitrogen escapes when the temperature rises to 47°C.