A sample of a gas occupies #1.40 * 10^3# mL at 25°C and 760 mmHg. What volume will it occupy at the same temperature and 380 mmHg?

Answer 1

#V = 2.8L#

On its initial state, the gas - occupies #1.4L# - is on a system of #298.15k#, and - #1 atm#
And on its second state, - #x L# - #298.15K# - #380/760 atm#
#(P_0V_0)/T_0 = (PV)/T#,

Find V.

#V = 2.8L#

Boyle's Law provides the solution: an inverse correlation factor exists between the state functions of pressure and volume when temperature is held constant.

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Answer 2

To find the volume of the gas at 380 mmHg, we can use Boyle's Law, which states that the pressure and volume of a gas are inversely proportional when temperature is constant.

( P_1 \times V_1 = P_2 \times V_2 )

Where: ( P_1 = 760 ) mmHg (initial pressure) ( V_1 = 1.40 \times 10^3 ) mL (initial volume) ( P_2 = 380 ) mmHg (final pressure) ( V_2 ) is the volume we want to find.

Solving for ( V_2 ):

( V_2 = \frac{{P_1 \times V_1}}{{P_2}} )

( V_2 = \frac{{760 \times 1.40 \times 10^3}}{{380}} )

( V_2 = 2.80 \times 10^3 ) mL

Therefore, the volume of the gas at 380 mmHg and 25°C would be ( 2.80 \times 10^3 ) mL.

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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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