Question

the work done by 2 moles of polyatomic gas initially at room temperature increases volume 8 times during adiabatic process will be
R=2cal /mol/K and room temperature =27K and ¥=4/3

Answer 1

Answer:

the work done by polyatomic gas will be 1800 cal

Answer 2

Correction : Room temperature = 27 °C or 300 K

Given :

Number of moles = 2

Initial temperature , t1 = 300 K

R = 2 cal/mol/k

Initial volume = V1

Final volume , V2 = 8 V1

γ = 4/3

To find :

The work done in the adiabatic process .

Solution :

Final temperature , t2 = t1 * ( V1 / V2 ) ^ (γ - 1)

                                t2 = 300 * ( 1/8 )^ ( 1/3 )

                                t2 = 300 * (1/2)

                                t2 = 150 K

Work done = [ R / (γ - 1) ] * [ T2 - T1 ]

                   = [ 2 / (1/3) ] * [ - 150 ]

                   = - 900 cal

The work done in the adiabatic process is - 900 cal .