The weight of `CH_(4)` in a 9L cylinder at `27^(@)C` temperature and 16 atm pressure is (R = 0.08 L atm `K^(-1) mol^(-1)` )

Options

A. 9.6g
B. 96.0 g
C. 4.8 g
D. 48.0 g

Explanation

To find the weight of CH₄ in a 9L cylinder at 27°C and 16 atm pressure, we can use the Ideal Gas Law, which is expressed as:

\[ PV = nRT \]

Where:
- \( P \) = pressure in atm
- \( V \) = volume in liters
- \( n \) = number of moles
- \( R \) = ideal gas constant
- \( T \) = temperature in Kelvin

### Step 1: Convert the temperature to Kelvin
The temperature in Celsius is given as 27°C. To convert this to Kelvin, we use the formula:

\[ T(K) = T(°C) + 273.15 \]

So,

\[ T = 27 + 273.15 = 300.15 \approx 300 \, K \]

### Step 2: Identify the values for the Ideal Gas Law
From the problem, we have:
- \( P = 16 \, atm \)
- \( V = 9 \, L \)
- \( R = 0.08 \, L \, atm \, K^{-1} \, mol^{-1} \)
- \( T = 300 \, K \)

### Step 3: Rearrange the Ideal Gas Law to solve for \( n \)
We can rearrange the Ideal Gas Law to find the number of moles \( n \):

\[ n = \frac{PV}{RT} \]

### Step 4: Substitute the values into the equation
Now, we substitute the known values into the equation:

\[ n = \frac{(16 \, atm)(9 \, L)}{(0.08 \, L \, atm \, K^{-1} \, mol^{-1})(300 \, K)} \]

### Step 5: Calculate \( n \)
Calculating the above expression:

\[ n = \frac{144}{24} = 6 \, moles \]

### Step 6: Calculate the mass of CH₄
The molecular mass of CH₄ (methane) is calculated as follows:

\[ \text{Molecular mass of CH₄} = 12 \, (C) + 4 \times 1 \, (H) = 16 \, g/mol \]

Now, we can find the mass of CH₄ using the number of moles:

\[ \text{Mass} = n \times \text{Molecular mass} \]

Substituting the values:

\[ \text{Mass} = 6 \, moles \times 16 \, g/mol = 96 \, g \]

### Final Answer
The weight of CH₄ in the cylinder is **96 grams**.

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