At 31 °C, the average translational kinetic energy of a gas molecule is K. If the temperature is now increased to 59 °C, the average translational kinetic energy of a molecule, in terms of K, will be closest to
Question
At 31 °C, the average translational kinetic energy of a gas molecule is K. If the temperature is now increased to 59 °C, the average translational kinetic energy of a molecule, in terms of K, will be closest to
Solution
The average translational kinetic energy of a gas molecule is directly proportional to the absolute temperature (in Kelvin). This relationship is described by the equation:
K1/K2 = T1/T2
where K1 and K2 are the kinetic energies at temperatures T1 and T2 respectively.
First, convert the temperatures from Celsius to Kelvin by adding 273.15 to each. This gives:
T1 = 31°C + 273.15 = 304.15 K T2 = 59°C + 273.15 = 332.15 K
Substitute these values into the equation:
K/ K2 = 304.15K / 332.15K
Solving for K2 gives:
K2 = K * (332.15 / 304.15)
So, the average translational kinetic energy of a molecule at 59 °C, in terms of K, is approximately 1.09K.
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