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 container if th

Question

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 container if the temperature remains constant?

1. 2.54 kPa

2. 157 kPa

3. 481 kPa

pls help ;(

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Trúc Chi 3 months 2021-07-29T14:20:09+00:00 1 Answers 5 views 0

Answers ( )

    0
    2021-07-29T14:22:06+00:00

    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 Boyle’s Law. 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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