A compressed cylinder of gas contains `1.50xx10^(3)` g of `N_(2)` gas at a pressure of `2.0xx10^(7)` Pa and a temperature of `17.1^(@)C`. What volume of gas has been released into the atmosphere if the final pressure in the cylinder is `1.80xx10^(5)` Pa ? Assume ideal behaviour and that the gas temperature is unchanged.

Options

A. 1264 L
B. 126 L
C. 12600 L
D. 45 L

Explanation

To solve the problem, we will follow these steps:

### Step 1: Convert the mass of nitrogen gas to moles.
Given:
- Mass of nitrogen gas (N₂) = \(1.50 \times 10^3 \, \text{g}\)
- Molar mass of nitrogen gas (N₂) = 28 g/mol

Using the formula:
\[ n = \frac{\text{mass}}{\text{molar mass}} \]

Calculating:
\[ n_1 = \frac{1.50 \times 10^3 \, \text{g}}{28 \, \text{g/mol}} = 53.57 \, \text{moles} \]

### Step 2: Use the ideal gas law to find the initial and final number of moles.
Using the ideal gas law, we know that:
\[ P_1V = n_1RT \quad \text{and} \quad P_2V = n_2RT \]

Since temperature (T) and volume (V) are constant, we can relate the pressures and number of moles:
\[ \frac{P_1}{P_2} = \frac{n_1}{n_2} \]

Given:
- Initial pressure \(P_1 = 2.0 \times 10^7 \, \text{Pa}\)
- Final pressure \(P_2 = 1.8 \times 10^5 \, \text{Pa}\)

Substituting the values:
\[ \frac{2.0 \times 10^7}{1.8 \times 10^5} = \frac{53.57}{n_2} \]

Calculating \(n_2\):
\[ n_2 = \frac{53.57 \times 1.8 \times 10^5}{2.0 \times 10^7} = 4.82 \, \text{moles} \]

### Step 3: Calculate the number of moles released.
The number of moles released (\(n_{\text{released}}\)) is given by:
\[ n_{\text{released}} = n_1 - n_2 \]

Calculating:
\[ n_{\text{released}} = 53.57 - 4.82 = 48.75 \, \text{moles} \]

### Step 4: Calculate the volume of gas released using the ideal gas law.
Using the ideal gas law:
\[ PV = nRT \]

We need to find the volume (V) of the released gas:
\[ V = \frac{nRT}{P} \]

Where:
- \(R = 8.314 \, \text{J/(mol K)}\) (or \(0.0821 \, \text{L atm/(mol K)}\))
- Temperature \(T = 17.1 + 273 = 290.1 \, \text{K}\)
- Pressure \(P = 1.8 \times 10^5 \, \text{Pa}\)

Calculating:
\[ V = \frac{48.75 \times 8.314 \times 290.1}{1.8 \times 10^5} \]

Calculating the volume:
\[ V \approx \frac{48.75 \times 8.314 \times 290.1}{1.8 \times 10^5} \approx 1264 \, \text{L} \]

### Final Answer:
The volume of gas that has been released into the atmosphere is approximately **1264 liters**.

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