Answer:
W= -28.37 KJ
Explanation:
Given
Gas is diatomic
We know that specific heat capacity ratio for diatomic gas ,Ī³ = 1.4
Pā = 1.4 atm ,Vā= 0.7 mĀ³
Pā= 2.1 atm
Lets take final volume = Vā
We know that for adiabatic process
[tex]P_1V_1^{\gamma}=P_2V_2^{\gamma}\\V_2=\left (\dfrac{P_1}{P_2} \right )^{\dfrac{1}{1.4}}V_1[/tex]
[tex]V_2=\left (\dfrac{1.4}{2.1}Ā \right )^{\dfrac{1}{1.4}}\times 0.7[/tex]
Vā = 0.52 mĀ³
The work done W
[tex]W=\dfrac{P_1V_1-P_2V_2}{\gamma-1}[/tex]
[tex]W=\dfrac{1.4\times 0.7-2.1\times 0.52}{1.4-1}[/tex]
W= -0.28 atm.mĀ³
This indicates that work is done on the gas.
W= - 0.28 atm.mĀ³
1 atm = 101325 Pa
W= -28371 Pa.mĀ³
W= -28.37 KJ
