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 given in the question.

### Step 1: Understand the Initial Conditions
We have a rigid container with 5 moles of \( H_2 \) gas. The initial conditions can be described using the Ideal Gas Law:
\[ PV = nRT \]
Where:
- \( P \) = initial pressure
- \( V \) = volume (constant since the container is rigid)
- \( n \) = number of moles (5 moles)
- \( R \) = universal gas constant
- \( T \) = initial temperature

### Step 2: Calculate the Initial Mass of Gas
The molar mass of \( H_2 \) is 2 g/mol. Therefore, the initial mass of the gas can be calculated as:
\[ \text{Mass} = n \times \text{Molar Mass} = 5 \, \text{moles} \times 2 \, \text{g/mol} = 10 \, \text{grams} \]

### Step 3: Determine the New Conditions After Gas Escapes
According to the problem:
- The pressure becomes half: \( P' = \frac{P}{2} \)
- The temperature becomes two-thirds: \( T' = \frac{2}{3}T \)

Let \( x \) be the number of moles of gas remaining after some gas has escaped.

### Step 4: Set Up the New Ideal Gas Law Equation
Using the new conditions, we can write the Ideal Gas Law for the remaining gas:
\[ P'V = n'RT' \]
Substituting the new values:
\[ \frac{P}{2}V = x \cdot R \cdot \left(\frac{2}{3}T\right) \]

### Step 5: Relate the Two Equations
Now, we can relate the initial and final states by dividing the two equations:
\[ \frac{PV}{nRT} \quad \text{(initial)} \quad \text{and} \quad \frac{\frac{P}{2}V}{xR\left(\frac{2}{3}T\right)} \quad \text{(final)} \]
This gives us:
\[ \frac{PV}{nRT} = \frac{\frac{P}{2}V}{xR\left(\frac{2}{3}T\right)} \]
Canceling out common terms:
\[ 1 = \frac{\frac{1}{2}}{x \cdot \frac{2}{3}} \]

### Step 6: Solve for \( x \)
Rearranging the equation:
\[ 1 = \frac{3}{4x} \implies 4x = 3 \implies x = \frac{3}{4} \]
Since we know the initial number of moles was 5, we can find the remaining moles:
\[ x = 5 - y \quad \text{(where \( y \) is the moles escaped)} \]

### Step 7: Calculate the Remaining Mass
The mass of the remaining gas can be calculated as:
\[ \text{Remaining mass} = x \cdot \text{Molar Mass} = \left(5 - y\right) \cdot 2 \]
Since we found \( x \) to be \( \frac{3}{4} \):
\[ \text{Remaining mass} = 7.5 \, \text{grams} \]

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

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