The enthalpy of vaporization of benzene, C6H6(l), is 33.9 kJ mol?1 at 298 K. How many liters of C6H6(g), measured at 298 K and 88.7 mmHg, are formed when 1.90 kJ of heat is absorbed by C6H6(l) at a constant temperature of 298 K?

Question

not sure what formulae to use, thanks!

M = Q/([heat of vaporization)
1.90 kJ/ 33.9 kJ/mol = 0.056 moles

Use the perfect gas law to covert that number of moles to liters of gas. Be sure to use the low pressure of 88.7 mm Hg

sweet dude got it

ihope thx

Answer 1

PV = nRT

0.056 moles * 0.0821 L atm K-1 mol-1 * 298 K = 11.3 L