Question

12.A thermally insulated diatomic ideal gas (N=2.5x1024) is adiabatically expanded from 0.10m to 0.20m. The initial pressure of the gas is 3.039x105 Pa. a. What is the internal energy change for the gas? b. Determine Q during this process. c. Calculate the work associated with this process.

Answer 1

-4 ca N - n = 123 V N=2.5810² Number of gas molecules. 12.5x1024 I moles MA 6.023x10 &ln=4.15 males IV = 0.10m3 Initial volume. Pi = 3.0398 10 Pal Inétical pressure. p= 0.20 m3 /> final valume. If initial temperature is Ti then, Pe Ve=nRT; T = Pe Vi = 3-039x105xool 4.15X8.314 IT = 880.79 K' Since, gas is deatomec so, adiabatic coefficient es = 7 r = 1 and specific heat capacity at constant volume os NR كارم

r-l process, Tur constant For adiabatic ... Tv =To Te ug f Vi IN To a 3-D f x Ti 0.10 ..20 x 880.79 K لا * T = ( 2 ) 0, 4 x 880.79.3 = 667.514K -:. Change in interncel energy is; AU=ncu AT = nx ERX (T-T) > AU= 4.15*$* 8.314x (667.514 – 880.79) Au=-18396.708 J D for adiabatic process no heat changes. So, Q=0J If work done by the gas es W then, from ist law of thermodynamics, W=Q-AU >> W=0- (-18396.708) J W=18396.708J =Wit4u