A vessel contains a mixture of `7 g` of nitrogen and `11 g` carbon dioxide at temperature ` T = 290 K`. If pressure of the mixure is `1 atm (= 1.01 xx 10^5 N//m^2)`, calculate its dencity `(R = 8.31 J//mol - K)`.

Explanation

To solve the problem, we need to calculate the density of the gas mixture using the ideal gas law and the given data. Here are the steps to find the solution:

### Step 1: Convert the mass of gases to kilograms
- Mass of nitrogen (N₂) = 7 g = \(7 \times 10^{-3}\) kg
- Mass of carbon dioxide (CO₂) = 11 g = \(11 \times 10^{-3}\) kg

### Step 2: Calculate the number of moles of each gas
- Molar mass of nitrogen (N₂) = 28 g/mol = \(28 \times 10^{-3}\) kg/mol
- Molar mass of carbon dioxide (CO₂) = 44 g/mol = \(44 \times 10^{-3}\) kg/mol

Using the formula for the number of moles:
\[ n = \frac{m}{M} \]
- Number of moles of nitrogen (\(n_{N2}\)):
\[ n_{N2} = \frac{7 \times 10^{-3}}{28 \times 10^{-3}} = \frac{7}{28} = 0.25 \text{ mol} \]
- Number of moles of carbon dioxide (\(n_{CO2}\)):
\[ n_{CO2} = \frac{11 \times 10^{-3}}{44 \times 10^{-3}} = \frac{11}{44} = 0.25 \text{ mol} \]

### Step 3: Calculate the total number of moles in the mixture
\[ n_{total} = n_{N2} + n_{CO2} = 0.25 + 0.25 = 0.5 \text{ mol} \]

### Step 4: Calculate the molar mass of the mixture
The molar mass of the mixture can be calculated using:
\[ M_{mix} = \frac{m_{N2} + m_{CO2}}{n_{total}} \]
\[ M_{mix} = \frac{7 \times 10^{-3} + 11 \times 10^{-3}}{0.5} = \frac{18 \times 10^{-3}}{0.5} = 36 \times 10^{-3} \text{ kg/mol} \]

### Step 5: Use the ideal gas law to find the density
The ideal gas law is given by:
\[ PV = nRT \]
From this, we can express density (\(\rho\)) as:
\[ \rho = \frac{PM}{RT} \]
Where:
- \(P = 1.01 \times 10^5 \text{ N/m}^2\)
- \(M = 36 \times 10^{-3} \text{ kg/mol}\)
- \(R = 8.31 \text{ J/(mol K)}\)
- \(T = 290 \text{ K}\)

Substituting the values:
\[ \rho = \frac{(1.01 \times 10^5) \times (36 \times 10^{-3})}{(8.31) \times (290)} \]

### Step 6: Calculate the density
Calculating the numerator:
\[ 1.01 \times 10^5 \times 36 \times 10^{-3} = 3636 \text{ N/m}^2 \]
Calculating the denominator:
\[ 8.31 \times 290 = 2410.9 \text{ J/(mol K)} \]
Now substituting these values into the density formula:
\[ \rho = \frac{3636}{2410.9} \approx 1.51 \text{ kg/m}^3 \]

### Final Answer
The density of the gas mixture is approximately \(1.51 \text{ kg/m}^3\).

---