Question

what is the average transaction kinetics energy of a molecule of an Ideal gas at a temperature of 29°c

Answer 1

To calculate the average translational kinetic energy of a molecule in an ideal gas, you can use the formula:

\[
\text{Average kinetic energy} (KE) = \frac{3}{2} k T
\]

where:
- \( k \) is the Boltzmann constant, approximately \( 1.38 \times 10^{-23} \, \text{J/K} \)
- \( T \) is the absolute temperature in Kelvin.

First, convert the temperature from degrees Celsius to Kelvin:

\[
T(K) = 29°C + 273.15 = 302.15 \, K
\]

Now, substitute \( T \) into the kinetic energy formula:

\[
KE = \frac{3}{2} (1.38 \times 10^{-23} \, \text{J/K}) (302.15 \, K)
\]

Calculating this gives:

\[
KE = \frac{3}{2} \times 1.38 \times 10^{-23} \times 302.15
\]

\[
KE = \frac{3}{2} \times 4.16457 \times 10^{-21} \, \text{J}
\]

\[
KE \approx 6.24686 \times 10^{-21} \, \text{J}
\]

Thus, the average translational kinetic energy of a molecule of an ideal gas at a temperature of 29°C is approximately \( 6.25 \times 10^{-21} \) J.