Question

At \( 27^\circ C \) kinetic energy of 4 g of \( H_2 \) is \( x \) J. What is the kinetic energy (in J) of 6.4 g of oxygen at \( 127^\circ C \)?

• A. \( \frac{x}{15} \)
• B. \( \frac{4x}{15} \)
• C. \( \frac{8x}{15} \)
• D. \( \frac{2x}{15} \)

Hint:

For kinetic energy problems in ideal gases, remember: \[ KE \propto nT \] where \( n \) is moles and \( T \) is temperature.

Answer 1

The Correct Option is C

Step 1: Kinetic Energy of an Ideal Gas The kinetic energy for an ideal gas is given by: \[ KE = \frac{3}{2} n R T \] where: - \( n \) is the number of moles, - \( R \) is the universal gas constant, - \( T \) is the temperature in Kelvin. Step 2: Computing \( n \) for \( H_2 \) Given 4 g of \( H_2 \): \[ n_{H_2} = \frac{4}{2} = 2 \text{ moles} \] Temperature conversion: \[ T_{H_2} = 27 + 273 = 300 K \] Step 3: Computing \( n \) for \( O_2 \) Given 6.4 g of \( O_2 \): \[ n_{O_2} = \frac{6.4}{32} = 0.2 \text{ moles} \] Temperature conversion: \[ T_{O_2} = 127 + 273 = 400 K \] Step 4: Ratio of Kinetic Energy Since \( KE \propto nT \): \[ \frac{KE_{O_2}}{KE_{H_2}} = \frac{0.2 \times 400}{2 \times 300} \] \[ = \frac{80}{600} = \frac{8}{15} \] Conclusion Thus, the correct answer is: \[ \frac{8x}{15} \]