Question

How much kinetic energy does a system containing 3 moles of an ideal gas at 300 K possess? What is the heat capacity at constant volume? How much heat would need to be transferred to the system to raise the temperature by 15°C?

Answer 1

1. K.E = 11.2239 kJ ≈ 11.224 kJ

2.
`C_(V) = 37.413 JK^(- 1)`

3.
`Q = 10.7749 kJ`

Solution:

Now, the kinetic energy of an ideal gas per mole is given by:

K.E =
`(3)/(2)mRT`

where

m = no. of moles = 3

R = Rydberg's constant = 8.314 J/mol.K

Temperature, T = 300 K

Therefore,

K.E =
`(3)/(2)* 3* 8.314* 300`

K.E = 11223.9 J = 11.2239 kJ ≈ 11.224 kJ

Now,

The heat capacity at constant volume is:

`C_(V) = (3)/(2)mR`

`C_(V) = (3)/(2)* 3* 8.314 = 37.413 JK^(- 1)`

Now,

Required heat transfer to raise the temperature by
`15^(\circ)` is:

`Q = C_(V)\Delta T`

`\Delta T = 15^(\circ) = 273 + 15 = 288 K`

`Q = 37.413* 288 = 10774.9 J = 10.7749 kJ`