Question

At 25 C and 1 atm, a balloon contains 0.1 mol of gas. How much work is involved in adding another 0.2 mol of gas to the balloon. What is the volume change in L?

Answer 1

PV = nRT

T = 25+273 = 298K

P = 1atm

n = 0.1moles

V = nRT/P

    = 0.1*0.0821*298/1 = 2.45L

PV = nRT

T = 25+273 = 298K

P = 1atm

n = 0.1+ 0.2 = 0.3moles

V = nRT/P

    = 0.3*0.0821*298/1 = 7.34L

chnage in vaolume = V = 7.34-2.45   = 4.89L

W   = PV

      = 1atm*4.89L   = 4.89L-atm

    = 4.89*101.3J                                       [ 1L-atm = 101.3J]

   = 495.36Joules >>>>answer