Question

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 1

Answer:

Work done will be 78.76 J

Explanation:

We have given initial volume of the gas $V_1=9.33\times 10^{-4}m^3$

Pressure is given by $P=1.013\times 10^5Pa$

Final volume $V_2=\frac{V_1}{6}=\frac{9.33\times 10^{-4}}{6}=1.555\times 10^{-4}m^3$

Change in volume $\Delta V=V_2-V_1=1.555\times 10^{-4}-9.33\times 10^{-4}=-7.775\times 10^{-4}m^3$

We know that work done is given by

$W=-Pdv=-1.013\times 10^5\times 7.775\times 10^{-4}=78.760J$

Work done will be 78.76 J