If #12 L# of a gas at room temperature exerts a pressure of #9 kPa# on its container, what pressure will the gas exert if the container's volume changes to #3 L#?

Answer 1

The pressure is #=36kPa#

Use #P_1V_1=P_2V_2# (Boyle's law) #P_1=9kPa# #V_1=12l# #V_2=3l#
#P_2=(P_1V_1)/V_2=(9*12)/3=36 kPa#
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Answer 2

Using the ideal gas law equation ( PV = nRT ), where ( P ) is pressure, ( V ) is volume, ( n ) is the number of moles of gas, ( R ) is the gas constant, and ( T ) is temperature, and assuming the number of moles and temperature remain constant, we can use the formula ( P_1V_1 = P_2V_2 ) to find the new pressure. Given that the initial volume is 12 L and the initial pressure is 9 kPa, and the final volume is 3 L, we can rearrange the formula to solve for the final pressure: ( P_2 = \frac{P_1V_1}{V_2} ).

Plugging in the values, we get ( P_2 = \frac{(9 , \text{kPa})(12 , \text{L})}{3 , \text{L}} = 36 , \text{kPa} ). Therefore, the gas will exert a pressure of 36 kPa when the container's volume changes to 3 L.

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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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