A 2L gas sample at 20\xb0C is cooled until it occupies a volume of 520 mL. What is the new temperature of the gas? Express your final answer in \xb0C



A 2L gas sample at 20°C is cooled until it occupies a volume of 520 mL. What is the new temperature of the gas? Express your final answer in °C​

Answer:

165°C

Explanation:

We can use here the combined gas law, relating the temperature, pressure, and volume of a gas with a constant quantity:

P₁V₁

T₁

P2V2

T2

If yore using the ideal-gas equation, which were NOT using here, you would have to convert each measurement into the appropriate units (L, K, atm, and mol). The only measurement that needs conversion here is temperature, from °C to K. (You will always convert temperature to Kelvin (absolute temperature) when using gas equations).

The Kelvin temperature is

K= 20°C + 273 = 293K

Lets rearrange the combined gas law to solve for the final temperature, T2:

T2

P2V2T1

P₁V₁

Plugging in known values, we can find the final temperature:

T2

(56.7kPa) (8.00L) (293K) (86.7kPa) (3.50L) 438K

Lastly, well convert back from K to °C:

438K 273 = 165°C



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