A sealed container holding 0.0255 L of an ideal gas at 0.983 atm and 67 ∘C is placed into a refrigerator and cooled to 45 ∘C with no change in volume. Calculate the final pressure of the gas.

Answer 1

Hello, Andrea,

Since no gas is added or subtracted from the container, we can use a simplified version of the gas laws: P1V1/T1 = P2V2/T2 (the moles, n, and the gas constant, R, both cancel and we can express the two states of the gas in one equation). T is always Kelvin, so add 273.15 to the temperatures to use Kelvin in gas law calcuations.

We want to find P2, the resulting pressure, so rearrange to solve for P2:

P2 = P1(V1/V2)(T2/T1)

Please note how I organized the right side of the equation. It is the product of the original pressure, P1, times the ratios of the (Initial/Final) Volumes and the (Final/Initial) pressures. This means that if the volume increases, the ratio of the two volumes will decrease, thereby reducing the pressure. If the temperature decreases, then the pressure would be lowered even further. Both scenarios make sense when you think about what's happening to the imprisioned gas. This also gives us a quick tool for predicting the outcome for P2. Look at the two ratios:

First of all V1/V2 is simply = 1, since the volume does not change. [I enjoy cancelations]. That leaves us (T2/T1), which is (318.2K/340.15 K), or around 90% of 1, about 0.9. That makes sense- we are reducing the temperature, so the gas molecules are not bouncing around as hard or fast. The pressure should drop by around 10%, to maybe 0.9, or so, just off the top of my head. We should expect a small drop.

Do the actual calculations. I find P2 to be 0.919 atm, which is close to what we just predicted without a calculator. That means we are ready to move ahead. The drop makes sense and the amount also was expected.

Bob