Collection of Solved Problems in Physics

Uknown Gas

Task number: 1282

• Hint – determination of initial temperature

  To determine the initial temperature T₁ of the unknown gas, we will use the formula for the total kinetic energy of one mole of gas \(\bar{E}_\mathrm{k}\).

• Formula for total kinetic energy of one mole

  The total kinetic energy of one mole \(\bar{E}_\mathrm{k}\) of any gas can be determined as follows:

  \[\bar{E}_\mathrm{k}=\frac{3}{2}nRT,\]

  where n is the amount of substance of the gas, T is the temperature and R is the molar gas constant.

• Hint – determination of the gas

  To determine the gas, you need to determine its molar mass M_m. To do this, you can use the relation for the most probable velocity v_p of gas molecules.

• The most probable velocity

  The most probable velocity v_p of the molecules of the heated gas is expressed by the relation

  \[v_\mathrm{p}=\sqrt{\frac{2RT_2}{M_\mathrm{m}}},\]

  where M_m is the molar mass of the gas, T₂ is the temperature of the heated gas and R is the molar gas constant.

• Numerical values

  \(\bar{E}_\mathrm{k}=2.5\,\mathrm{kJ}=2500\,\mathrm{J}\)	the total kinetic energy of one mole of monatomic gas
  n = 1 mole	the amount of substance of the monatomic gas
  ΔT = 300 K	the temperature difference
  vp = 642 ms−1	the most probable velocity of the gas molecules
  T1 = ?	the initial temperature

  ---

  From The Handbook of Chemistry and Physics:

  R = 8.31 Jmol−1K−1

  the molar gas constant

• Analysis

  We will determine the initial temperature from the formula decribing the total kinetic energy of one mole of gas, from which we can see that the energy is directly proportional to the thermodynamic temperature.

  To determine the gas we will evaluate the molar mass from the relation for the most probable velocity of gas molecules. Then we will compare the value with the values listed in The Handbook of Chemistry and Physics.

• Solution

  The total kinetic energy of one mole \(\bar{E}_\mathrm{k}\) of any gas relates with the thermodynamic temperature T through this equation

  \[\bar{E}_\mathrm{k}=\frac{3}{2}nRT,\]

  where n is the amount of the substance of the gas and R is the molar gas constant.

  This allows us to conduct a direct evaluation of the initial temperature T₁ of the unknown gas from the given values. It is true that:

  \[T_{1}=\frac{2\bar{E}_\mathrm{k}}{3nR}.\]

  The most probable velocity v_p of an ideal gas molecules is given by the relation

  \[v_\mathrm{p}=\sqrt{\frac{2RT_{2}}{M_m}},\]

  where M_m is the molar mass of the unknown gas and T₂ is the temperature after the gas is heated.

  Thereof we can directly express the molar mass of the unknown gas

  \[M_{\mathrm{m}}=\frac{2RT_{2}}{v^{2}_{\mathrm{p}}}.\]

  If we substitute the expression

  \[T_\mathrm{2}=T_\mathrm{1}+\mathrm{\Delta}T\]

  for the temperature T₂, where ΔT is the temperature difference, we obtain

  \[M_\mathrm{m}=\frac{2R(T_{\mathrm{1}}+\Delta T)}{v^{2}_{\mathrm{p}}}.\]

• Numerical solution

  \[T_{1}=\frac{2\bar{E}_\mathrm{k}}{3nR}=\frac{2\cdot{2500}}{3\cdot{1}\cdot{8.31}}\,\mathrm{K}\dot{=}200\,\mathrm{K}\]

   

  \[M_\mathrm{m}=\frac{2R(T_{\mathrm{1}}+\Delta T)}{v^{2}_{\mathrm{p}}}=\frac{2\cdot{8.31}\cdot(200+300)}{642^{2}}\,\mathrm{kg\,mol}^{-1}\] \[M_\mathrm{m}\dot{=}0.02\,\mathrm{kgmol}^{-1}=20\,\mathrm{g\,mol}^{-1}.\]

   

  Let us try to evaluate the most probable velocity at the initial temperature:

  \[v_{\mathrm{p}}^, = \sqrt{\frac{2RT_1}{M_\mathrm{m}}} = \sqrt{\frac{2RT_1}{\frac{2R(T_{1}+\Delta T)}{v^{2}_{\mathrm{p}}}}} = v_{\mathrm{p}}\,\sqrt{\frac{T_1}{T_{1}+\Delta T}}\] \[v_{\mathrm{p}}^, = 642\cdot\sqrt{\frac{200}{200+300}}\,\mathrm{m\,s^{-1}} \dot{=} 406\,\mathrm{m\,s^{-1}}\]

• Answer

  The molar mass of the unknown gas is about 20 g mol^-1, which corresponds to the value for neon. The initial temperature is approximately 200 K.