Question

5. Four moles of an ideal gas at 2.5 atm and 27° C are
compressed isothermally to half of its volume by an
external pressure of 3 atm. Calculate w,q and AU.​

Answer 1

Answer:

w = 5982.22 J

q = -5982.22 J

Δ U = 0

Explanation:

number of moles (n) = 4 moles

External pressure (P1) = 3 atm

Pressure of ideal gas (P) = 2.5 atm

Temperature of ideal gas = 27 degree Celsius = 27 + 273 = 300K

w = -P1 (V2 - V1)

Now initial volume ( V1) = nRT/P

V1 = 4 × 0.082 × 300/2.5

V1 = 39.36 L

Final volume (V2)= 39.36 L/2 = 19.68 L

Now

w = -P1 (V2 - V1)

w = -3 (19.68 L- 39.36 L)

w = 59.04 L atm

w = 59.04 × 101.325 = 5982.22 Joule

Since it is an isothermal process. So Δ U = 0

Δ U = q + w

0 = q +5982.22

q = -5982.22 J