Calculate the average kinetic energy of the N2 molecules in a sample of N2 gas at 273K and 546K.

I have been able to come up with the equation (KE)avg=3/2RT. I just want to know if R is equal to 0.08206 as it would be in other equations.

I believe R is 8.314. The 0.08206 has units of L*atm and you want joules.

To calculate the average kinetic energy (KE) of N2 molecules in a sample at different temperatures, you can use the equation you mentioned: (KE)avg = (3/2)RT.

In this equation, R represents the ideal gas constant, which has a value of 0.08206 L·atm/(mol·K). However, you should note that this value of R is specific to the units of liters, atmospheres, moles, and Kelvin. If you are using different units, such as joules, pascals, or kilopascals, you would need to use the appropriate value of R for those units.

Since you're using the equation at different temperatures, you'll need to calculate KE at both 273K and 546K. Let's do the calculations now:

For the sample at 273K:
(KE)avg = (3/2)RT
(KE)avg = (3/2)(0.08206 L·atm/(mol·K))(273K)
(KE)avg ≈ 332 J/mol

For the sample at 546K:
(KE)avg = (3/2)RT
(KE)avg = (3/2)(0.08206 L·atm/(mol·K))(546K)
(KE)avg ≈ 664 J/mol

Therefore, the average kinetic energy of the N2 molecules in the sample of N2 gas at 273K is approximately 332 J/mol, and at 546K is approximately 664 J/mol.

Remember to adjust the value of R based on the units you are using in your calculations.