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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Thermal equilibrium means that bodies in contact have the same temperature and no net heat flows between them. The topic distinguishes heat from temperature, uses the zeroth law of thermodynamics, and covers heat transfer, specific heat capacity, thermal expansion and the conditions for reaching equilibrium.
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