An ideal gas exerts a pressure of `1.52 MP`a when its temperature is `298.15 K` and its volume is `10^-2m^3`. (a) How many moles of gas are there ? (b) What is the mass density if the gas is molecular hydrogen ? ( c) What is the mass density if the gas is oxygen ?

Explanation

To solve the problem step by step, we will use the ideal gas 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 1: Calculate the number of moles of gas

Given:
- Pressure \( P = 1.52 \, \text{MPa} = 1.52 \times 10^6 \, \text{Pa} \)
- Volume \( V = 10^{-2} \, \text{m}^3 \)
- Temperature \( T = 298.15 \, \text{K} \)

Using the ideal gas equation, we can rearrange it to find the number of moles \( n \):

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

Substituting the values into the equation:

\[ n = \frac{(1.52 \times 10^6 \, \text{Pa}) \times (10^{-2} \, \text{m}^3)}{(8.314 \, \text{J/(mol K)}) \times (298.15 \, \text{K})} \]

Calculating the denominator:

\[ R \times T = 8.314 \times 298.15 \approx 2478.75 \, \text{J/mol} \]

Now substituting back into the equation for \( n \):

\[ n = \frac{(1.52 \times 10^6) \times (10^{-2})}{2478.75} \approx \frac{15200}{2478.75} \approx 6.13 \, \text{moles} \]

### Step 2: Calculate the mass density if the gas is molecular hydrogen

The molecular mass of hydrogen \( H_2 \) is approximately \( 2 \, \text{g/mol} = 0.002 \, \text{kg/mol} \).

Using the formula for density derived from the ideal gas equation:

\[ \text{Density} (\rho) = \frac{P \times \text{Molecular Mass}}{RT} \]

Substituting the values for hydrogen:

\[ \rho = \frac{(1.52 \times 10^6) \times (0.002)}{(8.314) \times (298.15)} \]

Calculating:

\[ \rho = \frac{3040}{2478.75} \approx 1.22 \, \text{kg/m}^3 \]

### Step 3: Calculate the mass density if the gas is oxygen

The molecular mass of oxygen \( O_2 \) is approximately \( 32 \, \text{g/mol} = 0.032 \, \text{kg/mol} \).

Using the same density formula:

\[ \rho = \frac{P \times \text{Molecular Mass}}{RT} \]

Substituting the values for oxygen:

\[ \rho = \frac{(1.52 \times 10^6) \times (0.032)}{(8.314) \times (298.15)} \]

Calculating:

\[ \rho = \frac{48640}{2478.75} \approx 19.61 \, \text{kg/m}^3 \]

### Final Answers:
(a) The number of moles of gas is approximately \( 6.13 \, \text{moles} \).
(b) The mass density of hydrogen is approximately \( 1.22 \, \text{kg/m}^3 \).
(c) The mass density of oxygen is approximately \( 19.61 \, \text{kg/m}^3 \).