Question

Calculate the work done by the system in an irreversible (single step) adiabatic expansion of 2 mole of a polyatomic gas (`gamma`= 4/3) from 300 K and pressure 10 atm to 1 atm : ? (a) -227 R (b) -205 R (c) -405 R (d) None of these

Answer 1

`d`

Explanation:

Work done in case of adiabatic process is expressed as `-nC_v delta T`

Where, `delta T` is the change in temperature.

Now,for adiabatic process we know,

`T^(gamma)/P^(gamma-1)`= constant

So,using this we can find the final temperature,

`((T_2)/(T_1))^(gamma) = ((P_2)/(P_1))^(gamma-1)`

Given, `T_1=300K,P_2=1 atm,P_1=10 atm,gamma=4/3`

So, `T_2=168.70 K`

hence, `delta T=(300-168.70)=131.3 K`

Given, `gamma =4/3=1+2/f` (where, `f` is the degrees of freedom)

So,we get, `f=6`

Now, `C_v=(f/2)R=3R`

So,work done by the system is

`-nC_v delta T=-2*3R*131.3=-787.8R`