<h2>
Answer:</h2>
810.93
<h2>
Step-by-step explanation:</h2>
Let the pressure be given by P and the volume be V.
Since pressure is inversely proportional to volume, we can write;
P ∝ ![\frac{1}{V}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7BV%7D)
=> P =
-------------(i)
Where;
c = constant of proportionality.
<em>When the volume of the gas is 2 cubic feet, pressure is 1000 pounds per square foot.</em>
V = 2 ft³
P = 1000lb/ft²
Substitute these values into equation (i) as follows;
1000 = ![\frac{c}{2}](https://tex.z-dn.net/?f=%5Cfrac%7Bc%7D%7B2%7D)
=> c = 2 x 1000
=> c = 2000 <em>lbft</em>
<em />
Substituting this value of c back into equation (i) gives
P = ![\frac{2000}{V}](https://tex.z-dn.net/?f=%5Cfrac%7B2000%7D%7BV%7D)
This is the general equation for the relation between the pressure and the volume of the given gas.
To calculate the work done <em>W</em> by the gas, we use the formula
![W = \int\limits^{V_1}_{V_0} {P} \, dV](https://tex.z-dn.net/?f=W%20%3D%20%5Cint%5Climits%5E%7BV_1%7D_%7BV_0%7D%20%7BP%7D%20%5C%2C%20dV)
Where;
V₁ = final volume of the gas = 3ft³
V₀ = initial volume of the gas = 2ft³
<em>Substitute P = </em>
<em>, V₁ = 3ft³ and V₀ = 2ft³</em>
![W = \int\limits^{3}_{2} {\frac{2000}{V} } \, dV](https://tex.z-dn.net/?f=W%20%3D%20%5Cint%5Climits%5E%7B3%7D_%7B2%7D%20%7B%5Cfrac%7B2000%7D%7BV%7D%20%7D%20%5C%2C%20dV)
<em>Integrate</em>
W = 2000ln[V]³₂
W = 2000(In[3] - ln[2])
W = 2000(0.405465108)
W = 810.93016
W = 810.93 [to 2 decimal places]
Therefore, the work done by the gas for the given pressure and volume is 810.93