Question

A quantity of gas with an initial volume of 5 cubic feet and a pressure of 1700 pounds per square foot expands to a volume of 9 cubic feet. Find the work done by the gas for the given volume and pressure. Round your answer to two decimal places. Assume the temperature of the gas in this process remain constant.

Answer 1

Answer:

Work done by the gas for the given volume and pressure $= 4996.18$ pounds foot

Explanation:

Given

Pressure applied $= 1700$ pounds per square foot

Initial Volume $= 5$ cubic feet

Final Volume $= 9$ cubic feet

As we know pressure is inversely proportional to V

$P = \frac{k}{V}$

where k is the proportionality constant

V is the volume and

P is the pressure

Work done

$\int\limits^{V_2}_{V_1} {P} \, dV$

 $\int\limits^{V_2}_{V_1} {\frac{k}{V} } \, dV\\= \int\limits^{V_2}_{V_1} {\frac{1700 * 5}{V} } \, dV\\= 8500* \int\limits^{V_2}_{V_1} {\frac{1}{V} } \, dV$

Integrating the above equation, we get-

$8500 ln \frac{V_2}{V_1} \\8500 * 2.303 * \frac{9}{5} \\= 4996.18$

Work done by the gas for the given volume and pressure $= 4996.18$ pounds foot