A sample of sulfur hexafluoride gas occupies a volume of 9.10 L at 198ÁC. Assuming that the pressure remains constant, what temperature (in ÁC) is needed to reduce the volume to 2.50 L?

Respuesta :

  As we know that
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

Temperature -143.6°C is needed to reduce the volume to 2.50 L

Further explanation  

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  

V1: V2 = n1: n2  

• 2. Boyle's Law  

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

p1.V1 = p2.V2  

• 3. Charles's Law  

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

V1 / T1 = V2 / T2  

• 4. Gay Lussac's Law  

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

P1 / T1 = P2 / T2  

• 5. Law of Boyle-Gay-Lussac  

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

P1.V1 / T1 = P2.V2 / T2  

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 :

[tex]\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[/tex]

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