The rapid depletion of fossil fuels has inspired extensive research in the area of alternative and renewale energy sources. Of these, hydroden is the most Contemplated fuel of the future . However cost effective production and hazard free storage are major issuses is using `H_(2)` . A cylinder contains hydrogen at a pressure of 80MPa at `25^(@)C` , Assuming ideal behaviour , Calculate the density of hydrogen in the cylinder in `Kg//m^(3)` .

Answer

To solve the problem of calculating the density of hydrogen gas in a cylinder under the given conditions, we can follow these steps:

### Step 1: Understand the Ideal Gas Law
The ideal gas law is given by the equation:
\[ PV = nRT \]
Where:
- \( P \) = pressure (in Pascals)
- \( V \) = volume (in cubic meters)
- \( n \) = number of moles
- \( R \) = universal gas constant (\( 8.314 \, \text{J/(mol·K)} \))
- \( T \) = temperature (in Kelvin)

### Step 2: Convert Given Values to Appropriate Units
1. **Pressure**: The pressure is given as \( 80 \, \text{MPa} \). To convert this to Pascals:
\[ P = 80 \, \text{MPa} = 80 \times 10^6 \, \text{Pa} \]

2. **Temperature**: The temperature is given as \( 25^\circ C \). To convert this to Kelvin:
\[ T = 25 + 273 = 298 \, \text{K} \]

3. **Molecular Mass of Hydrogen**: The molecular mass of hydrogen (\( H_2 \)) is \( 2 \, \text{g/mol} \). To convert this to kilograms:
\[ M = 2 \, \text{g/mol} = 2 \times 10^{-3} \, \text{kg/mol} \]

### Step 3: Rearranging the Ideal Gas Law to Find Density
Density (\( \rho \)) is defined as mass per unit volume. From the ideal gas law, we can express density as:
\[ \rho = \frac{mass}{volume} = \frac{nM}{V} \]
Using \( n = \frac{PV}{RT} \), we can substitute:
\[ \rho = \frac{PM}{RT} \]

### Step 4: Substitute Values into the Density Formula
Now we can substitute the values we have into the density formula:
\[ \rho = \frac{(80 \times 10^6 \, \text{Pa}) \times (2 \times 10^{-3} \, \text{kg/mol})}{(8.314 \, \text{J/(mol·K)}) \times (298 \, \text{K})} \]

### Step 5: Calculate the Density
Now we perform the calculation:
1. Calculate the numerator:
\[ 80 \times 10^6 \times 2 \times 10^{-3} = 160000 \, \text{kg·Pa} \]

2. Calculate the denominator:
\[ 8.314 \times 298 = 2477.572 \, \text{J/(mol·K)} \]

3. Now divide the numerator by the denominator:
\[ \rho = \frac{160000}{2477.572} \approx 64.57 \, \text{kg/m}^3 \]

### Final Answer
The density of hydrogen in the cylinder is approximately:
\[ \rho \approx 64.57 \, \text{kg/m}^3 \]

---