Question

The kinetic energy of translation of the molecules in 50 g of CO\(_2\) gas at 17°C is:

• A. 4102.8 J
• B. 4205.5 J
• C. 3986.3 J
• D. 3582.7 J

Hint:

Use the ideal gas law relations to compute kinetic energy in gas molecules.

Answer 1

The Correct Option is A

• The translational kinetic energy is given by: \[ (KE)_{{translational}} = \left[ \frac{3}{2} kT \right] \times \text{No. of molecules} \]
• Number of molecules: \[ \frac{50}{44} \times 6.023 \times 10^{23} \]
• Calculation: \[ (KE)_{{translational}} = 4108.644 \, \text{J} \]

Answer 2

Step 1: Understanding the problem.
We are asked to find the total kinetic energy of translation of the molecules in 50 g of CO₂ gas at 17°C.

Step 2: Recall the formula for total translational kinetic energy.
For an ideal gas, the total translational kinetic energy is given by:
\[ E = \frac{3}{2}nRT \] where:
\( n \) = number of moles of gas,
\( R \) = universal gas constant = 8.314 J mol⁻¹ K⁻¹,
\( T \) = temperature in Kelvin.

Step 3: Convert the given quantities.
Given mass of CO₂ = 50 g.
Molar mass of CO₂ = 44 g mol⁻¹.
\[ n = \frac{50}{44} = 1.136 \, \text{mol}. \] Temperature = 17°C = 17 + 273 = 290 K.

Step 4: Substitute in the formula.
\[ E = \frac{3}{2} \times n \times R \times T \] \[ E = \frac{3}{2} \times 1.136 \times 8.314 \times 290 \] \[ E = 1.5 \times 1.136 \times 8.314 \times 290 = 4102.8 \, \text{J}. \]

Step 5: Final Answer.
\[ \boxed{E = 4102.8 \, \text{J}} \]