Question

Just need the answer to this one

1. How much heat is released at constant pressure if a 20.0-L tank containing 59.0 atm of hydrogen sulfide gas condenses at its boiling point of -60.0°C? The enthalpy of vaporization of hydrogen sulfide is 18.7 kJ/mol at-60.00. (R A) 4.48 x 106 J B) 2.77 x 102 C) 7.81 x 101 D) 1.26 x 106 J E) 1.87×104 J 0.0821 L·atm/(K·mol))

Answer 1

Given conditions are Pressure P=59 atm

Temperature T=-60 C=213 K

Volume V=20 L

Let n= no. of moles of the gas

From ideal gas law we have

PV=nRT

So n= PV/RT

so n= (59x20)/(0.0821x213)

n=67.47 moles

As Enthalpy of vaporization is 18.7 kJ/mol al 213 K

so heat released for 67.47 moles will be = 18.7 x 67.47 =1260 kJ

So option D is correct answer.