Question

n moles of gaseous nitrogen are cooled off from a Temperature T=300 K to just below its liquid temperature of 77 K with a specific enthalpy change ΔH1= -1440 cal/mol. This amount of nitrogen can be liquefied by extracting some extra heat. The specific heat of vaporization of liquid nitrogen equals ΔH2= 1336 cal/mol. Calculate the total amount of heat to be extracted from n=1.3 moles of gaseous nitrogen in this process.

Answer 1

Given the moles of gaseous nitrogen, n = 1.3 mol

Given the specific enthalpy change to reduce temperature from 300K to 77K = - 1440 cal/mol

Hence the specific enthalpy change required by 1.3 mol of N2 to reduce temperature from 300K to 77K, H1

= - 1440 cal/mol x 1.3 mol = - 1872 cal

Given the specific heat of vaporization of liquid nitrogen = 1336 cal/mol

Hence specific heat of vaporization of 1.3 mol of liquid nitrogen = 1336 cal/mol x 1.3 mol = 1737 cal.

Now to convert gaseous nitrogen to liquid we need to extract an amount of heat which is equal and opposite to heat of vaporization.

Hence H2 = - 1737 cal

Hence total amount of heat need to be extracted = H1 + H2 =  - 1872 cal -  1737 cal

= - 3609 cal (answer)