Question

150 grams of C₂H₆ an ideal gas has an initial pressure of 9120 mmHg and a temperature of 300 K. At a constant temperature and moles, the gas changes to a final pressure is 2280 mmHg.

a) Calculate the initial and final volumes (L)

b) Calculate the work done (in kJ) for the gas volume change if it is carried out against a constant external pressure of 6 atm. (1 L atm = 101.325 J)

c) Using answer 5b, is this an exothermic or endothermic process if the heat is equal to +11000 J?

Answer 1

Moles of C_{​​​​​​2}​​​​​H_{​​​​6} = given mass / molar mass = 150 g / (30 g/mol) = 5 mol

P_{​​​​​​Initial} = 9120 mmHg = 9120/760 atm = 12 atm

P_{​​​​​​Final} = 2280 mmHg = 2280/760 atm = 3 atm

T = 300 K

a. pV = nRT => V = nRT/p

V_{​​​​​​Initial} = (5 mol)*(0.0821 L-atm/mol-K)*(300 K) / 12 atm = 10.26 L

V_{​​​​​Final} = (5 mol)*(0.0821 L-atm/mol-K)*(300 K) / 3 atm = 41.05 L

b. Work done in Isothermal process = nRT ln(V_{​​​​​​Final} / V_{​​​​​​Initial} ) = (5 mol)*(0.0821 L-atm/mol-K)*(300 K)*(ln(41.05/10.26)) = 170.72 L-atm = 170.72 × 101.325 J = 17.3 kJ

c. This is exothermic reaction.