Consider a cylinder initially filled with 9.33 10-4 m3 of ideal gas at atmospheric pressure. An external force is applied to slowly compress the gas at constant temperature to 1/6 of its initial volume. Calculate the work that is done. Note that atmospheric pressure is 1.013 105 Pa.

Answer:Work done will be 78.76 JExplanation: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