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3 years ago

635 mL of a gas is at a pressure of 8.00 atm. What is the volume of the gas at standard pressure (STP)?

Chemistry

1 answer:

Sveta_85 [38]3 years ago

5 0

The new volume at standard temperature and pressure is 5.08 L.

Explanation:

As per the kinetic theory of gases, the volume occupied by gas molecules will be inversely proportional to the pressure of the gas molecules. This is termed as Boyle's law.

So, pressure∝ $\frac{1}{volume}$

Thus, if two pressure and two volumes are given then,

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

Now, we known the values of P₁ = 8 atm, V₁ = 635 mL, P₂ = 1 atm and V₂ we have to determine. We are considering P₂ = 1 atm, because we have to determine V₂ at standard temperature and pressure. And standard pressure is 1 atm.

$8*635*10^{-3} = 1 *V{2} \\\\V_{2} = \frac{8*635*10^{-3} }{1} = 5.08 L$

Thus, the new volume at standard temperature and pressure is 5.08 L.

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3 years ago

How many moles of gas X are present if the gas has a volume of 2dm³ at room temperature and pressure? Give your answer to 2 deci

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Answer:

Approximately $0.08\; \rm mol$ , assuming that this gas is an ideal gas.

Explanation:

Look up the standard room temperature and pressure: $25\; \rm ^{\circ}C$ and $P = 101.325 \; \rm kPa$ .

The question states that the volume of this gas is $V = 2\; \rm dm^{3}$ .

Convert the unit of all three measures to standard units:

$\begin{aligned} T &= 25\; \rm ^{\circ}C \\ &= (25 + 273.15)\; \rm K \\ &= 293.15\; \rm K\end{aligned}$ .

$\begin{aligned}P &= 101.325\; \rm kPa \\ &= 101.325 \; \rm kPa \times \frac{10^{3}\; \rm Pa}{1\; \rm kPa} \\ &= 1.01325 \times 10^{5}\; \rm Pa\end{aligned}$ .

$\begin{aligned}V &= 2\; \rm dm^{3} \\ &= 2 \; \rm dm^{3} \times \frac{1\; \rm m^{3}}{10^{3}\; \rm dm^{3}} \\ &= 2 \times 10^{-3}\; \rm m^{3}\end{aligned}$ .

Look up the ideal gas constant in the corresponding units: $R \approx 8.31\; \rm m^{3}\cdot Pa \cdot mol^{-1} \cdot K^{-1}$ .

Let $n$ denote the number of moles of this gas in that $V = 2\; \rm dm^{3}$ . By the ideal gas law, if this gas is an ideal gas, then the following equation would hold:

$P \cdot V = n \cdot R \cdot T$ .

Rearrange this equation and solve for $n$ :

$\begin{aligned}n &= \frac{P \cdot V}{R \cdot T} \\ &\approx \frac{1.01325 \times 10^{5}\; {\rm Pa} \times 2 \times 10^{-3}\; {\rm m^{3}}}{8.31 \; {\rm m^{3} \cdot Pa \cdot mol^{-1} \cdot K^{-1}} \times 293.15\; {\rm K}} \\ &\approx 0.08\; \rm mol\end{aligned}$ .

In other words, there is approximately $2\; \rm mol$ of this gas in that $V = 2\; \rm dm^{3}$ .

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