$\frac 2{3}$
$\frac 5{7}$
When an ideal diatomic gas is heated at constant pressure, the total heat energy supplied \( Q \) is partly used to increase the internal energy of the gas and partly to do work against the external pressure. The fraction of heat energy that increases the internal energy can be calculated by considering the specific heat capacities.
The relationship for heat supplied at constant pressure \( Q \) is given by: \[ Q = n C_P \Delta T \] Where: - \( n \) is the number of moles, - \( C_P \) is the specific heat capacity at constant pressure, - \( \Delta T \) is the temperature change. For an ideal diatomic gas, the specific heat capacity at constant pressure \( C_P \) is related to the specific heat at constant volume \( C_V \) by: \[ C_P = C_V + R \] For a diatomic gas, \( C_V = \frac{5}{2} R \) (because the degrees of freedom for a diatomic gas are 5), and: \[ C_P = \frac{7}{2} R \] The heat energy that increases the internal energy of the gas is the part associated with the change in internal energy, which is given by: \[ \Delta U = n C_V \Delta T \] The fraction of the total heat energy supplied that increases the internal energy is: \[ \text{Fraction} = \frac{\Delta U}{Q} = \frac{n C_V \Delta T}{n C_P \Delta T} = \frac{C_V}{C_P} \] Substituting the values for \( C_V \) and \( C_P \): \[ \text{Fraction} = \frac{\frac{5}{2} R}{\frac{7}{2} R} = \frac{5}{7} \] Thus, the fraction of heat energy that increases the internal energy of the gas is \( \frac{5}{7} \).
Correct Answer: (E) \( \frac{5}{7} \)
The standard heat of formation, in kcal/mol, of $Ba^{2+}$ is:
Given: Standard heat of formation of SO₄²⁻(aq) = -216 kcal/mol, standard heat of crystallization of BaSO₄(s) = -4.5 kcal/mol, standard heat of formation of BaSO₄(s) = -349 kcal/mol.