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3 years ago

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A sample of Ar gas has a volume of 6.30 L with an unknown pressure. The gas has a volume of 8.57 L when the pressure is 2.74 atm

, with no change in temperature or amount of gas. What was the initial pressure, in atmospheres, of the gas?

1 answer:

ExtremeBDS [4]3 years ago

8 0

Answer:

3.73 atm

Explanation:

Since there is no change in temperature, we use Boyle's law.

P₁V₁ = P₂V₂ where P₁ = initial pressure of Ar gas = unknown, P₂ = final pressure of Ar gas = 2.74 atm, V₁ = initial volume of Ar gas = 6.30 L and V₂ = final volume of Ar gas = 8.57 L

P₁ = P₂V₂/V₁ = 2.74 atm × 8.57 L/6.30 L = 3.73 atm

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A gas has a volume of 1.75L at -23°C and 150.0 kPa. At what temperature would the gas occupy 1.30L at 210.0 kPa?

Nastasia [14]

Answer:

At -13 $^{0}\textrm{C}$ , the gas would occupy 1.30L at 210.0 kPa.

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$\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}$

where $P_{1}$ and $P_{2}$ are initial and final pressure respectively.

$V_{1}$ and $V_{2}$ are initial and final volume respectively.

$T_{1}$ and $T_{2}$ are initial and final temperature in kelvin scale respectively.

Here $P_{1}=150.0kPa$ , $V_{1}=1.75L$ , $T_{1}=(273-23)K=250K$ , $P_{2}=210.0kPa$ and $V_{2}=1.30L$

Hence $T_{2}=\frac{P_{2}V_{2}T_{1}}{P_{1}V_{1}}$

$\Rightarrow T_{2}=\frac{(210.0kPa)\times (1.30L)\times (250K)}{(150.0kPa)\times (1.75L)}$

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