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0.016 mol of a diatomic gas, with initial temperature 18 ∘C , are compressed from 1300...

0.016 mol of a diatomic gas, with initial temperature 18 ∘C , are compressed from 1300 cm3 to 550 cm3 in a process in which pV2=constant

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Answer #1

n=0.016

T initial=291K

Vi=0.0013 m3

Vf=0.00055 m3

initial pressure Pi= nRT/V

= 29.77 kPa =0.293atm

The detailed behavior of the pressure and volume depends upon the specific heats of the gas:

Constant pressure =  J/mol K
Constant volume =  J/mol K
The specific heats determine the ratio, =

1.4 for diatomic gas

Pf=2.8 atm

substitute all values

Tf= 1187K

Finally,

heat is added during the process Q or temperature increased by the factor 1187K/291K =4

Answer is 4

Note: Reference SerwayJewett chapter 21-Kinetic Threoy of Gases

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