At 46°C a sample of ammonia gas exerts a pressure of 5.3 atm. What is the pressure when the volume of the gas is reduced to one-fourth (0.250) of the original value at the same temperature?

Hello there!

As you may know, pressure and volume have an inverse relationship. This means that as pressure increases, volume decreases (and as volume increases, pressure decreases).

In this problem, the volume of the gas is reduced to 0.250 its original value. This means that the pressure should be higher than it was originally.

We can use Boyle's Law to solve this problem, seeing that temperature is kept constant.

P1V1 = P2V2

Let's rearrange this equation to isolate the variable we need, P2.

P1V1

  = P2

 V2

That's better! Now that we have that down, let's plug in the numbers we were given.

(5.3atm) * (V1)

= P2

0.25(V1)

We can get rid of V1, as the number would be equal to 1 (anything divided by itself is equal to 1).

5.3atm

= 21.2 atm

0.25L

Now, while the number is significantly higher, this is quite reasonable.

Let's plug it in and see if we get what we should:

5.3(V1) = (0.25V1)(21.2)

0.25V1

21.2 atm = 21.2 atm

Seems right to me!

Feel free to comment if you need any more help. :)

i believe it is 40–40–20%

explanation:

the density(ratio between mass and volume as m/v) decreases from top to bottom in periodic table group as the size increases and nuclear charge decreases. but in periods density increases from left to right as atom size decreases and nuclear charge increases.