Question

What is the average translational KE of 5 moles of gas molecules at 300 K?

a.
2.1 x 103 J

b.
1.87 x 104 J

c.
1.78 x 105 J

d.
1.68 x 106 J

Answer 1

The average translational kinetic energy of 5 moles of gas molecules at 300 K is 1.87 x 10^4 J, calculated using the ideal gas law.

Step-by-step explanation:

The average translational kinetic energy (KEtrans) of one mole of ideal gas molecules at a temperature T is given by the equation KEtrans = (3/2) RT, where R is the ideal gas constant (8.314 J/mol·K) and T is the temperature in kelvins. For n moles of gas, the total translational kinetic energy would be KEtotal = n × KEtrans. Substituting the given values into this equation (n = 5 moles, T = 300 K) gives KEtotal = 5 × (3/2) × 8.314 J/mol·K × 300 K.

Performing the calculation, KEtotal = 5 × (3/2) × 8.314 × 300 = 5 × 3.957 × 103 J = 1.87 × 104 J. Therefore, the correct answer is (b) 1.87 × 104 J, which represents the average translational kinetic energy of 5 moles of gas molecules at 300 K.

Answer 2

What is the average translational kinetic energy of molecules in an ideal gas at 37°C? The average translational energy of a molecule is given by the equipartition theorem as, E = 3kT 2 where k is the Boltzmann constant and T is the absolute temperature.

Step-by-step explanation:

The average translational energy of a molecule is given by the equipartition theorem as, E = 3kT 2 where k is the Boltzmann constant and T is the absolute temperature.