A vessel has 6gm of oxygen at pressure P and temperature 400 K.A small hole is made in it so that oxygen leaks out. How much oxygen leaks out if the final pressure is `(P)/(2)` and temperauture 300 K ?

Options

A. A. 5 g
B. B. 4 g
C. C. 2 g
D. D. 3 g

Explanation

To solve the problem, we will use the Ideal Gas Law, which states that:

\[ PV = nRT \]

where:
- \( P \) = pressure,
- \( V \) = volume,
- \( n \) = number of moles,
- \( R \) = universal gas constant,
- \( T \) = temperature in Kelvin.

### Step 1: Determine Initial Conditions

Initially, we have:
- Mass of oxygen, \( m_1 = 6 \) g
- Initial pressure, \( P_1 = P \)
- Initial temperature, \( T_1 = 400 \) K

To find the initial number of moles \( n_1 \), we use the molar mass of oxygen, which is approximately \( 32 \) g/mol:

\[ n_1 = \frac{m_1}{M} = \frac{6 \, \text{g}}{32 \, \text{g/mol}} = \frac{3}{16} \, \text{mol} \]

### Step 2: Set Up the Ideal Gas Law for Initial Conditions

Using the Ideal Gas Law for the initial state:

\[ PV = n_1RT_1 \]

Substituting \( n_1 \):

\[ PV = \left(\frac{3}{16}\right) R (400) \]

### Step 3: Determine Final Conditions

After some oxygen leaks out, we have:
- Final pressure, \( P_2 = \frac{P}{2} \)
- Final temperature, \( T_2 = 300 \) K
- Let the mass of oxygen remaining be \( m_2 \).

The number of moles after leakage \( n_2 \) can be expressed as:

\[ n_2 = \frac{m_2}{M} \]

### Step 4: Set Up the Ideal Gas Law for Final Conditions

Using the Ideal Gas Law for the final state:

\[ \frac{P}{2} V = n_2 R T_2 \]

Substituting \( n_2 \):

\[ \frac{P}{2} V = \left(\frac{m_2}{M}\right) R (300) \]

### Step 5: Relate Initial and Final States

Since the volume \( V \) is the same in both cases, we can equate the two expressions:

\[ PV = \left(\frac{3}{16}\right) R (400) \quad \text{(1)} \]
\[ \frac{P}{2} V = \left(\frac{m_2}{M}\right) R (300) \quad \text{(2)} \]

From (1), we can express \( V \):

\[ V = \frac{\left(\frac{3}{16}\right) R (400)}{P} \]

Substituting \( V \) into (2):

\[ \frac{P}{2} \cdot \frac{\left(\frac{3}{16}\right) R (400)}{P} = \left(\frac{m_2}{M}\right) R (300) \]

### Step 6: Simplify and Solve for \( m_2 \)

Cancel \( R \) and \( P \):

\[ \frac{3}{32} (400) = \left(\frac{m_2}{M}\right) (300) \]

Now, simplify:

\[ \frac{3 \cdot 400}{32 \cdot 300} = \frac{m_2}{M} \]

Calculating the left side:

\[ \frac{1200}{9600} = \frac{m_2}{M} \implies \frac{1}{8} = \frac{m_2}{M} \]

Thus:

\[ m_2 = \frac{M}{8} \]

### Step 7: Calculate the Mass of Oxygen Remaining

Substituting \( M = 32 \) g/mol:

\[ m_2 = \frac{32}{8} = 4 \, \text{g} \]

### Step 8: Find the Mass of Oxygen Leaked

The mass of oxygen that leaked out is:

\[ \text{Mass leaked} = m_1 - m_2 = 6 \, \text{g} - 4 \, \text{g} = 2 \, \text{g} \]

### Final Answer

The amount of oxygen that leaks out is **2 grams**.

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