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.

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Doug Chapman 7/10/08