Change in the Internal Energy of Oxygen

Task number: 1286

What is the internal energy difference of oxygen with the mass of 10 g, when it is heated from 10 °C to 60 °C, given that the heating process is being carried out

a) at constant volume,

b) at constant pressure,

c) by adiabatic compression?

Consider the gas to be ideal.

  • Hint 1

    What does the internal energy of ideal gas depend on?

  • Hint 2

    Write the relation between the internal energy difference and the temperature difference.

  • Hint 3

    The molar heat capacity of oxygen at constant volume is

    \[C_V=\frac{5}{2}R.\]
  • Analysis

    The internal energy of an ideal gas is a state function depending only on the temperature. The result will therefore be the same for all three cases.

    We will express the internal energy difference as a product of the amount of substance of oxygen, the molar heat capacity of oxygen at constant volume and the difference between the initial and final temperatures.

    The unknown amount of substance can be evaluated by dividing the mass by the molar mass of oxygen.

  • Numerical values

    m = 100 g = 0.100 kg the mass of oxygen
    t1 = 10 °C the initial temperature of oxygen
    t2 = 60 °C the final temperature of oxygen
    ΔU = ? the internal energy difference of oxygen

    From The Handbook of Chemistry and Physics:

    R = 8.31 JK−1mol−1 the molar gas constant
    Mm = 32 g mol−1 the molar mass of oxygen O2
  • Solution

    The internal energy of an ideal gas is a state function depending only on the temperature. This means that in all three considered cases, the result will be the same, because it depends only on the initial and the final temperature of the gas, and not on the way the gas was heated to the final temperature.

    The molar heat capacity of oxygen CV at constant volume is

    \[C_V=\frac{5}{2}R\]

    and the amount of substance n is given by the formula

    \[n=\frac{m}{M_m},\]

    Therefore the internal energy difference ΔU can be expressed as follows:

    \[\Delta U=nC_V\left(T_2-T_1\right)= \frac{m}{M_m}\,\frac{5}{2}R\,\left(T_2-T_1\right).\]

    If we substitute the temperature difference t2 − t1 for the thermodynamic temperature difference T2 − T1 and adjust the fraction, we get

    \[\Delta U=\frac{5mR}{2M_m}\left(t_2-t_1\right).\]
  • Numerical insertion

    \[\Delta U=\frac{5mR}{2M_m}\,\left(t_2-t_1\right)= \frac{5\cdot{ 0.100}\cdot{ 8.31}}{2\cdot{0.032}}\cdot \left(60-10\right)\,\mathrm{J}\dot{=}3250\,\mathrm{J}\]
  • Answer

    The change in the internal energy in all considered cases is approximately 3250 J.

Difficulty level: Level 2 – Upper secondary level
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