Calculate kinetic energy of 5 moles of nitrogen at 27°c

General
1

calculate kinetic energy of 5 moles of nitrogen at 27°c

2

Calculate kinetic energy of 5 moles of nitrogen at 27°C

Answer:
The kinetic energy of a gas can be calculated using the formula:
[ KE = \frac{3}{2}RT ]
Where:

  • KE is the kinetic energy,
  • R is the ideal gas constant, and
  • T is the temperature in Kelvin.

Given that we have 5 moles of nitrogen gas at 27°C, we first need to convert the temperature to Kelvin by adding 273 to the Celsius temperature:
[ T = 27°C + 273 = 300K ]

Now, substituting the values into the formula:
[ KE = \frac{3}{2} \times 8.314 , \text{J/mol·K} \times 300K \times 5 ]
[ KE = \frac{3}{2} \times 8.314 \times 300 \times 5 ]
[ KE = 3 \times 8.314 \times 1500 ]
[ KE = 37362.6 , \text{Joules} ]

Therefore, the kinetic energy of 5 moles of nitrogen gas at 27°C is 37362.6 Joules.