A `2.00 mL` volume container contains `50 mg` of gas at a pressure of `100 kPa`. The mass of each gas particle is `8.0 xx 10^-26 kg`. Find the average translational kinetic energy of each particle

Explanation

To find the average translational kinetic energy of each gas particle in the given scenario, we can follow these steps:

### Step 1: Convert the mass of gas to grams
Given mass of gas = 50 mg = \(50 \times 10^{-3}\) g = 0.050 g.

### Step 2: Calculate the molar mass of the gas
The mass of each gas particle is given as \(8.0 \times 10^{-26}\) kg.
To find the molar mass (M) in grams per mole, we can use Avogadro's number (\(N_A = 6.022 \times 10^{23}\) particles/mol):

\[ M = \text{mass of one particle} \times N_A = (8.0 \times 10^{-26} \, \text{kg}) \times (6.022 \times 10^{23} \, \text{particles/mol}) \]

Converting kg to g:
\[ M = (8.0 \times 10^{-26} \times 10^3 \, \text{g}) \times (6.022 \times 10^{23}) = 4.818 \, \text{g/mol} \]

### Step 3: Calculate the number of moles of gas
Using the formula:
\[ n = \frac{\text{mass}}{\text{molar mass}} = \frac{0.050 \, \text{g}}{4.818 \, \text{g/mol}} \approx 0.0104 \, \text{mol} \]

### Step 4: Calculate the temperature using the ideal gas law
Using the ideal gas equation:
\[ PV = nRT \]
Where:
- \(P = 100 \, \text{kPa} = 100,000 \, \text{Pa}\)
- \(V = 2 \, \text{mL} = 2 \times 10^{-6} \, \text{m}^3\)
- \(R = 8.314 \, \text{J/(mol K)}\)
- \(n = 0.0104 \, \text{mol}\)

Rearranging for temperature \(T\):
\[ T = \frac{PV}{nR} = \frac{(100,000 \, \text{Pa}) \times (2 \times 10^{-6} \, \text{m}^3)}{(0.0104 \, \text{mol}) \times (8.314 \, \text{J/(mol K)})} \]

Calculating:
\[ T \approx \frac{0.2}{0.0865} \approx 23.1 \, \text{K} \]

### Step 5: Calculate the average translational kinetic energy
The average translational kinetic energy \(E_k\) of each particle is given by:
\[ E_k = \frac{3}{2} k T \]
Where \(k\) is the Boltzmann constant \(k = 1.38 \times 10^{-23} \, \text{J/K}\).

Substituting the values:
\[ E_k = \frac{3}{2} \times (1.38 \times 10^{-23} \, \text{J/K}) \times (23.1 \, \text{K}) \approx 4.78 \times 10^{-22} \, \text{J} \]

### Final Answer
The average translational kinetic energy of each particle is approximately \(4.78 \times 10^{-22} \, \text{J}\).

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