Question

Find the average translational kinetic energy per molecule if one mole of the gas is contained in a volume .00123 metrecube at a pressure 200000Nper metresquare

Answer 1

Answer on Question #64815 - Physics – Molecular Physics | Thermodynamisc

Question: Find the average translational kinetic energy per molecule if one mole of the gas is contained in a volume .00123 metrecube at a pressure 200000N per metresquare

Solution: Accord to the basic equation of kinetic theory of gases $P = \frac{2}{3} \frac{N}{V} E$ where $P$ is pressure (in Pa), $N$ is a molecule's number, $V$ is volume (in $\mathrm{m}^3$) and $E$ is average translational kinetic energy (in J). The molecule's number is product of Avogadro's number and mole's number ($N = N_A * \nu$). So, $E = \frac{3PV}{2N_A \nu}$.

Given: $P = 200000 \, \text{N/m}^2 = 2 \times 10^5 \, \text{Pa}$; $V = 0.00123 \, \text{m}^3 = 1.23 \times 10^{-3} \, \text{m}^3$; $\nu = 1$ mole. $N_A = 6.02 \times 10^{23}$.

Calculate: $E = \frac{3 \times 2 \times 10^{5} \, \text{Pa} \times 1.23 \times 10^{-3} \, \text{m}^{3}}{2 \times 6.02 \times 10^{23} \times 1} = \frac{7.38 \times 10^{2} \, \text{N} \times \text{m}}{12.04 \times 10^{23}} = 0.613 \times 10^{-21} \, \text{N} \times \text{m} = 6.13 \times 10^{-22} \, \text{J}$

Answer: The average translational kinetic energy per molecule in the system is $6.13 \times 10^{-22}$ Joules.

Note: The kinetic theory of gases works only for ideal gases.

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