## 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 V_{1} and pressure P_{1}; we then
decrease the volume to 1/4 of the original volume V_{1} to a new volume V_{2} and pressure P_{2}. Since the
number of moles of gas (as well as the temperature) are constant,
we have P_{1}V_{1}=P_{2}V_{2}. Now, the
volume was decreased to 1/4V_{1} to give the final volume V_{2}; therefore,
V_{2} = 1/4V_{1}. Substituting this into
P_{1}V_{1}=P_{2}V_{2} and solving for
P_{2}, we find that P_{2} = 4P_{1}. 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 P_{1}/T_{1}=P_{2}/T_{2}. Since the temperature
was halved from state 1 to state 2, we have T_{2} = T_{1}/2. Substituting
and solving for P_{2}, we find that P_{2} = P_{1}/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 P_{1}/n_{1}=P_{2}/n_{2}. Notice that n_{2} = 1/2n_{1},
and substitute. You should see that P_{2} = 1/2P_{1}. At constant T and V, pressure and
quantity (moles) are directly proportional.

If you have any questions or comments, reply to chapman@sou.edu

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Doug Chapman chapman@sou.edu 7/10/08