A rigid container contains 5 mole `H_(2)` gas at some pressure and temperature. The gas has been allowed to escape by simple process from the container due to which pressure of the gas becomes half of its initial pressure and temperature become `(2//3)^(rd)` of its initial. The mass of gas remaining is :

Options

A. 7.5 g
B. 1.5 g
C. 2.5 g
D. 3.5 g

Explanation

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 start with 5 moles of \( H_2 \) gas in a rigid container. The initial pressure is \( P \) and the initial temperature is \( T \).

### Step 2: Write the Ideal Gas Law for Initial Conditions
Using the ideal gas law:
\[ PV = nRT \]
For the initial conditions:
\[ PV = 5RT \quad \text{(1)} \]

### Step 3: Understand the Final Conditions
After some gas escapes, the final pressure \( P' \) is half of the initial pressure:
\[ P' = \frac{P}{2} \]
The final temperature \( T' \) is two-thirds of the initial temperature:
\[ T' = \frac{2}{3}T \]

### Step 4: Write the Ideal Gas Law for Final Conditions
For the final conditions, we have:
\[ P'V = n'RT' \]
Substituting the expressions for \( P' \) and \( T' \):
\[ \frac{P}{2}V = n'R\left(\frac{2}{3}T\right) \quad \text{(2)} \]

### Step 5: Divide the Two Equations
Now, we divide equation (2) by equation (1):
\[ \frac{\frac{P}{2}V}{PV} = \frac{n'R\left(\frac{2}{3}T\right)}{5RT} \]
This simplifies to:
\[ \frac{1}{2} = \frac{n' \cdot \frac{2}{3}}{5} \]

### Step 6: Solve for \( n' \)
Rearranging gives us:
\[ \frac{1}{2} = \frac{2n'}{15} \]
Cross-multiplying:
\[ 15 = 4n' \]
Thus:
\[ n' = \frac{15}{4} \text{ moles} \]

### Step 7: Calculate the Mass of Remaining Gas
The mass of the remaining gas can be calculated using the formula:
\[ \text{mass} = n' \times \text{molar mass} \]
The molar mass of \( H_2 \) is 2 g/mol, so:
\[ \text{mass} = \frac{15}{4} \times 2 = \frac{30}{4} = 7.5 \text{ grams} \]

### Final Answer
The mass of the gas remaining in the container is **7.5 grams**.

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