Answer:
Work done will be 78.76 J
Explanation:
We have given initial volume of the gas [tex]V_1=9.33\times 10^{-4}m^3[/tex]
Pressure is given by [tex]P=1.013\times 10^5Pa[/tex]
Final volume [tex]V_2=\frac{V_1}{6}=\frac{9.33\times 10^{-4}}{6}=1.555\times 10^{-4}m^3[/tex]
Change in volume [tex]\Delta V=V_2-V_1=1.555\times 10^{-4}-9.33\times 10^{-4}=-7.775\times 10^{-4}m^3[/tex]
We know that work done is given by
[tex]W=-Pdv=-1.013\times 10^5\times 7.775\times 10^{-4}=78.760J[/tex]
Work done will be 78.76 J
