Question

A monatomic ideal gas undergoes an isothermal expansion at 300 k, as the volume increased from 0.010 m3 to 0.040 m3. the final pressure is 130 kpa. what is the change in the internal (thermal) energy of the gas during this process? (r = 8.31 j/mol ⢠k)

Answer 1

To determine the change in the internal energy of the system, we use the first law of thermodynamics which expresses the change in internal energy. It is expressed as:

ΔU = Q + W

where ΔU is the change in total energy
           Q is the heat energy
          W is work

It is said that the system undergoes an isothermal process so the value of Q would be zero. The change in internal energy would be equal to work. In thermodynamics, work is expressed as:

Work = ∫ - PdV

Since P is not constant, we need to express it in terms of V and substitute it to the equation. We use PV = nRT

P = nRT/V

Work = ∫ - (nRT/V)dV

Integrating from V1 to V2, we will have:

Work = nRT ln V2/V1

ΔU = nRT ln (V2/V1) = 1 (8.314) (300) ln (0.040 / 0.010)
ΔU = 34457.40 J