Chem hw help please (: A sample of an ideal gas in a cylinder of volume 2.99 L at 298 K and 2.77 atm expands to 7.73 L by two different pathways. PathA is an isothermal, reversible expansion. Path B has two steps. In the first step, the gas is cooled at constant volume to 1.29 atm.In the second step, the gas is heated and allowed to expand against a constant external pressure of 1.29 atm until the final volumeis 7.73 L.Calculate the work for path A.W =JCalculate the work for path B.W =J

First, we have to find the moles of the gas inside the cylinder. Initial state: V = 2.99 L T = 298 K P = 2.77 atm moles of gas, n = ?? Using ideal gas equation n = PVRT⇒n = 2.77 atm * 2.99 L0.08206 L.atm.mol-1.K-1 * 298 K⇒n = 0.33869 mol#Path-A: Reversible isothermal expansion: Initial volume, V1 = 2.99 L Final volume, V2 = 7.73 L The formula for work done in reversible isothermal expansion is: w = -nRT*lnV2V1⇒w = - 0.33869 mol * 8.314 J.mol-1.K-1 * 298 K * ln7.73 L2.99 L⇒w =- 797 J (Answer)# Path-B: #First step: Since volume is constant in 1st step, work done is equal to 0 (w1 = 0). #2nd step: The gas is heated against a constant external pressure, Pext = 1.29 atm Initial volume, V1 = 2.99 L Final volume, V2 = 7.73 L Change in volume, ∆V = V2 - V1 = 7.73 L - 2.99 L = 4.74 L Formula for work done against constant external pressure is: w2 = - Pext * ∆V ⇒w2 = - 1.29 atm * 4.74 L⇒w2 =- 6.1146 L.atm⇒w2 =- 6.1146 L.atm * 101.325 J1 L.atm⇒w2 =- 620. J Total work, w = w1+w2 => w = 0+ (- 620.)J => w = - 620. J (Answer)