Question

Method 1 of DeltaU Temperature of two moles of a monoatomic gas is increased by 600 K in a given process. Find change in internal energy of the gas.

Answer 1

Answer:

hello mate

pls.. follow me

Answer 2

Change in internal energy = 216.5J

Explanation:

according to kinetic theory of gases

change in internal energy of a ideal gas is directly proportional to temperature change.

As mentioned above temperature change = 600K

gas is mono-atomic

no. of moles = 2

  Change in internal energy = (number of moles ) (specific heat at constant volume) (temperature change)

  ΔU = n$c_{v}$ΔT

$c_{v} =\frac{3}{2R}$  (for mono - atomic gases)

change in internal energy = 2×600 ×  $\frac{3}{2R}$

                                          =1800÷ 8.314

                                       ΔU  = 216.5J