What is the average translational kinetic energy of molecules in a gas at temperature 27°C?
Correct answer: C. 6.21 x 10^-21 J
- A. 3.23 x 10^-21 J
- B. 4.11 x 10^-21 J
- C. 6.21 x 10^-21 J
- D. 7.71. x 107-21 J
- E. 9.11 x 10^7-21 J
Explanation
To calculate the average translational kinetic energy of molecules in a gas at a given temperature, we can use the equation:K.E. = (3/2) kTWhere K.E. is the average translational kinetic energy, k is the Boltzmann constant (approximately 1.38 x 10^-23 J/K), and T is the temperature in Kelvin.Given that the temperature is 27°C, we need to convert it to Kelvin by adding 273.15:T = 27°C + 273.15 = 300.15 KSubstituting the values into the equation:K.E. = (3/2) * (1.38 x 10^-23 J/K) * (300.15 K)= 6.21 x 10-21 JTherefore, the average translational kinetic energy of molecules in the gas at a temperature of 27°C is approximately 6.21 x 10-21 J.
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