If #18 L# of a gas at room temperature exerts a pressure of #35 kPa# on its container, what pressure will the gas exert if the container's volume changes to #14 L#?

Answer 1

The new pressure is #45 kPa#

Let's start off with identifying our known and unknown variables. The first volume we have is #18# L, the first pressure #35kPa# and the second volume is #14L#. Our only unknown is the second pressure.

Boyle's Law, which states that pressure and volume have an inverse relationship as long as temperature and mole count are constant, can be used to find the solution.

The equation we use is #P_1V_1=P_2V_2# where the numbers 1 and 2 represent the first and second conditions. All we have to do is rearrange the equation to solve for the pressure.
We do this by dividing both sides by #V_2# in order to get #P_2# by itself like so: #P_2=(P_1xxV_1)/V_2#
Now all we do is plug and chug! #P_2=(35\kPa xx 18\ cancel"L")/(14\cancel"L")# = #45# #kPa#
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Answer 2

( P_1V_1 = P_2V_2 )
( (35 , \text{kPa})(18 , \text{L}) = P_2(14 , \text{L}) )
( P_2 = \frac{(35 , \text{kPa})(18 , \text{L})}{14 , \text{L}} )
( P_2 \approx 45.0 , \text{kPa} )

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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