Problem 10.25

Assume that you have a cylinder with a movable piston. What would happen to the gas pressure inside the cylinder if you do the following?

(a)Decrease the volume to one-fourth the original volume while holding the temperature constant

Consult the figure on P. 401 in your text, or better yet, sketch out a piston/cylinder setup yourself. The variables of interest here (P, V, T, n) are related by the ideal gas law:

PV / T = nR = constant

Now, we let the initial state have a volume V₁ and pressure P₁; we then decrease the volume to 1/4 of the original volume V₁ to a new volume V₂ and pressure P₂. Since the number of moles of gas (as well as the temperature) are constant, we have P₁V₁=P₂V₂. Now, the volume was decreased to 1/4V₁ to give the final volume V₂; therefore, V₂ = 1/4V₁. Substituting this into P₁V₁=P₂V₂ and solving for P₂, we find that P₂ = 4P₁. For a decrease in volume by a factor of 4, the pressure increases by a factor of 4. This should make sense since, at constant T and n, pressure and volume are inversely propotional.

(b)Reduce the Kelvin temperature to half its original value while holding the volume constant

For constant volume and moles, we write the ideal gas law as P/T = nR/V = constant. Using the same idea as above, when the temperature is halved at constant volume, we have P₁/T₁=P₂/T₂. Since the temperature was halved from state 1 to state 2, we have T₂ = T₁/2. Substituting and solving for P₂, we find that P₂ = P₁/2. Thus, at constant n and V, pressure and temperature are directly proportional, and lowering the temperature by a factor of 2 lowers the pressure by a factor of 2 also.

(c) Reduce the amount of gas to half while keeping the volume and temperature constant

Same deal here: at constant T, V we write P/n = RT/V. Set up the equation as P₁/n₁=P₂/n₂. Notice that n₂ = 1/2n₁, and substitute. You should see that P₂ = 1/2P₁. At constant T and V, pressure and quantity (moles) are directly proportional.