If 7/5 L of a gas at room temperature exerts a pressure of 6 kPa on its container, what pressure will the gas exert if the container's volume changes to 2/3 L?

1 Answer
Aug 1, 2016

The gas will exert a pressure of 63/5 kPa.

Explanation:

Let's begin by identifying our known and unknown variables.

The first volume we have is 7/5 L, the first pressure is 6kPa and the second volume is 2/3L. Our only unknown is the second pressure.

We can obtain the answer using Boyle's Law:
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The letters i and f represent the initial and final conditions. All we have to do is rearrange the equation to solve for the final pressure.

We do this by dividing both sides by V_f in order to get P_f by itself like so:
P_f=(P_ixxV_i)/V_f

Now all we do is plug in the values and we're done!

P_f=(6\kPa xx 7/5\ cancel"L")/(2/3\cancel"L") = 63/5kPa