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Tcecarenko [31]
3 years ago
6

A sample of sulfur hexafluoride gas occupies a volume of 9.10 L at 198ÁC. Assuming that the pressure remains constant, what temp

erature (in ÁC) is needed to reduce the volume to 2.50 L?
Chemistry
2 answers:
alexandr1967 [171]3 years ago
5 0
  As we know that
<span>V1/T1 = V2/T2 
V1 = 9.10 L 
T1 = 471 K 
V2 = 2.50 L 
T2 = 2.5 x 471 / 9.10 = 129.3 K 
T2 = 129.3 - 273 =
 -143.6 deg Celsiu
hope it helps</span>
cupoosta [38]3 years ago
5 0

Temperature <u>-143.6°C</u> is needed to reduce the volume to 2.50 L

<h3>Further explanation  </h3>

There are several gas equations in various processes:  

• 1. Avogadro's hypothesis  

In the same temperature and pressure, in the same volume conditions, the gas contains the same number of molecules  

So it applies: the ratio of gas volume will be equal to the ratio of gas moles  

<h3>V1: V2 = n1: n2  </h3>

• 2. Boyle's Law  

At a fixed temperature, the gas volume is inversely proportional to the pressure applied  

<h3>p1.V1 = p2.V2  </h3>

• 3. Charles's Law  

When the gas pressure is kept constant, the gas volume is proportional to the temperature  

<h3>V1 / T1 = V2 / T2  </h3>

• 4. Gay Lussac's Law  

When the volume is not changed, the gas pressure in the tube is proportional to its absolute temperature  

<h3>P1 / T1 = P2 / T2  </h3>

• 5. Law of Boyle-Gay-Lussac  

Combined with Boyle's law and Gay Lussac's law  

<h3>P1.V1 / T1 = P2.V2 / T2  </h3>

P1 = initial gas pressure (N / m2 or Pa)  

V1 = initial gas volume (m3)  

P2 = gas end pressure  

V2 = the final volume of gas  

T1 = initial gas temperature (K)  

T2 = gas end temperature  

In the problem, the conditions that are set constant are Pressure, so we use Charles' Law

V1 / T1 = V2 / T2

V1 = 9.1 L

T1 = 198 °C+273 = 471 K

V2 = 2.5 L

Then :

\displaystyle \frac{9.1}{471}=\frac{2.5}{T2}\\\\T2=\frac{471\times 2.5}{9.1}\\\\T2=129.4 K\\\\=129.4-273=-143.6^oC

<h3> Learn more </h3>

a description of Charles’s law

brainly.com/question/5056208

Charles's law

brainly.com/question/9510865

State Boyle's, Charles's, and Gay-Lussac's laws

brainly.com/question/980439

Keywords : Charles Law, temperature, pressure, vollume, gas, sulfur hexafluoride

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<em></em>

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Answer: There are 1.469 \times 10^{23} molecules present in 7.62 L of CH_4 at 87.5^{o}C and 722 torr.

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Converting torr into atm as follows.

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Substitute the values into above formula as follows.

PV = nRT\\0.95 atm \times 7.62 L = n \times 0.0821 L atm/mol K \times 360.5 K\\n = \frac{0.95 atm \times 7.62 L}{0.0821 L atm/mol K \times 360.5 K}\\= \frac{7.239}{29.59705}\\= 0.244 mol

According to the mole concept, 1 mole of every substance contains 6.022 \times 10^{23} atoms. Hence, number of atoms or molecules present in 0.244 mol are calculated as follows.

0.244 mol \times 6.022 \times 10^{23}\\= 1.469 \times 10^{23}

Thus, we can conclude that there are 1.469 \times 10^{23} molecules present in 7.62 L of CH_4 at 87.5^{o}C and 722 torr.

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