What is the average translational kinetic energy of one mole Of gas molecule in an ideal gas at 37degree celsius
T=273+37=310 K
KE=3νRT/2 =3•1•8.31•310/2 =3864 J
To calculate the average translational kinetic energy (KE) of one mole of gas molecules in an ideal gas at a specific temperature (37 degrees Celsius), we can use the equation:
KE = (3/2) * N * k * T
Where:
KE is the average translational kinetic energy
N is the Avogadro's number (6.022 x 10^23 mol^(-1))
k is the Boltzmann constant (1.38 x 10^(-23) J/K)
T is the temperature in Kelvin (37°C + 273.15 = 310.15 K)
Now, let's substitute the values into the equation:
KE = (3/2) * (6.022 x 10^23) * (1.38 x 10^(-23)) * 310.15
By simplifying the equation, we get:
KE ≈ 1.20 x 10^(-21) J
So, the average translational kinetic energy of one mole of gas molecules in an ideal gas at 37 degrees Celsius is approximately 1.20 x 10^(-21) Joules.