Let us write the classical ideal gas law for neon and oxygen respectively: "p_1 V_1 = \\frac{m_1}{M_1} R T_1", "p_2 V_2 = \\frac{m_2}{M_2} R T_2". Dividing two equations of state, obtain: "\\frac{p_1 V_1}{p_2 V_2} = \\frac{m_1}{m_2}\\frac{M_2}{M_1}\\frac{T_1}{T_2}". According to the given conditions, the tank has a fixed volume, hence "V_1 = V_2", and the pressure and temperature is the same, hence "p_1 = p_2", "T_1 = T_2". Hence, two equations of state simplify to "1 = \\frac{m_1}{m_2}\\frac{M_1}{M_2}" , from where "m_2 = m_1 \\frac{M_2}{M_1}" . The molar masses of neon and oxygen are "M_1 = 10 \\frac{g}{mol}" and "M_2 = 16 \\frac{g}{mol}" respectively, hence the mass of the oxygen is "m_2 = 2 kg \\cdot \\frac{16 \\frac{g}{mol}}{10 \\frac{g}{mol}} = 3.2 kg".
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