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Enter your answer in the provided box. A sample of sulfur hexafluoride gas occupies 9.53 L at 215°C. Assuming that the pressure remains constant, what temperature (in °C) is needed to reduce the volume to 4.36 L? Report your answer to the proper number of significant figures.

User Bluebadge
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1 Answer

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Step-by-step explanation:

According to ideal gas equation, the product of pressure and volume equals to the product of number of moles, gas constant and temperature.

Mathematically, PV = nRT

Since, it is given that pressure is constant and Charle's law states that at constant pressure volume of an ideal gas is directly proportional to the absolute temperature.

Therefore,
V \propto T


(V)/(T) = constant

So, number of moles will also be constant then. Hence, the formula will be as follows.


(V_(1))/(T_(1)) =
(V_(2))/(T_(2))

As it is given that
V_(1) is 9.53 L,
V_(2) is 4.36 L,
T_(1) is
215^(o)C that is also equal to (215 + 273) K = 488 K.

Now, putting these values into the formula as follows.


(V_(1))/(T_(1)) =
(V_(2))/(T_(2))


(9.53 L)/(488 K) =
(4.36 L)/(T_(2))


T_(2) = 223.26 K

Temperature in degree celsius will be (223.26 K - 273 K) =
-49.73^(o)C

Thus, we can conclude that temperature (in °C) is needed to reduce the volume to 4.36 L is
-49.73^(o)C.

User MichK
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