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shusha [124]
3 years ago
10

A gas with a volume of 20.0l at a pressure of 275 kpa is allowed to expand to a volume of 35.0l. what is the pressure in the con

tainer if the temperature remains constant?
1. 2.54 kPa

2. 157 kPa

3. 481 kPa

pls help ;(
Chemistry
1 answer:
sladkih [1.3K]3 years ago
8 0

Answer:

\boxed {\boxed {\sf 2. \ 157 \ kPa}}

Explanation:

Since temperature remains constant, the only variables that change are volume and pressure. Therefore, we are using <u>Boyle's Law.</u> This states that the pressure is inversely proportional to the volume. The formula is:

P_1V_1=P_2V_2

We know the gas starts with a volume of 20.0 liters at a pressure of 275 kPa. We can substitute these values into the left side of the formula.

275 \ kPa *20.0 \ L=P_2V_2

We know the gas expands to a volume of 35.0 Liters, but we do not know the pressure.

275 \ kPa *20.0 \ L=P_2* 35.0 \ L

Since we are solving for the new pressure, we must isolate the variable P₂. It is being multiplied by 35.0 Liters and the inverse of multiplication is division. Divide both sides by 35.0 L.

\frac {275 \ kPa * 20.0 \ L}{ 35.0 \ L}= \frac{P_2*35.0 \ L}{35.0 \ L}

\frac {275 \ kPa * 20.0 \ L}{ 35.0 \ L}=P_2

The units of liters cancel.

\frac {275 \ kPa * 20.0 }{ 35.0 }=P_2

\frac {5500}{35.0} \ kPa= P_2

157.142857 \  kPa=P_2

The original measurements of pressure and volume have 3 significant figures, so our answer must have the same. For the number we calculated, that is the ones place.

The 1 in the tenths place (157.142857) tells us to leave the 7 in the ones place.

157 \ kPa= P_2

If the gas expanded to a volume of 35.0 liters while the temperature remained constant, the pressure in container was approximately 157 kilopascals.

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