Question

1) When you pour liquid nitrogen into a jack-o-lantern, it looks cool. What is the ASvap for nitrogen given that its AHvap is 5.57 kJ/mol at its boiling temperature of -195.81 °C?

Answer 1

Ans) ∆S_vap =72.02 kJ/mol.K

Formual to be used :

T_b = ∆H_vap /∆S_vap

T_b is boiling temperature

∆H_vap is enthalpy of vaporization

∆S_vap is entropy of vaporization

Given : ∆H_{vap =} 5.57 kJ/mol

1kJ = 1000 J

5.57 kJ/mol = 5.57*1000 kJ/mol = 5570 kJ/mol

∆H_{vap =} 5570 kJ/mol

T_b = -195.81 °C

Temperature (K) = 273.15 + Temperature(°C)

Temperature (K) =273.15 +(-195.81 °C)

Temperature (K) =273.15 -195.81

Temperature (K) =77.34 K

T_b =77.34 K

Putting all the values in formula

T_b = ∆H_vap /∆S_vap

77.34 K =( 5570 kJ/mol)/∆S_vap

∆S_vap = ( 5570 kJ/mol)/77.34 K

∆S_vap =( 5570/77.34 ) kJ/mol.K

∆S_vap =72.02 kJ/mol.K