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lions [1.4K]
3 years ago
13

A gas sample has a temperature of 22c with an unknown volume. The same gas has a volume of 456 mL when the temperature is 86c wi

th no change in the pressure or amount of gas. What was the initial volume in millimeters oh the gas
Chemistry
1 answer:
tester [92]3 years ago
3 0

Answer:

V₁  = 374.71  mL

Explanation:

Given data:

Initial volume of gas= ?

Initial temperature = 22°C

Final temperature = 86°C

Final volume = 456 mL

Solution:

Initial temperature = 22°C (22+273 = 295 k)

Final temperature = 86°C (86+273 = 359 k)

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₁ = V₂T₁ /T₂

V₁  = 456 mL × 295 K / 359 k

V₁  = 134520 mL.K /  359 k

V₁  = 374.71  mL

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F a balloon containing 1000 L of gas 50 Celsius and 101.3kpa rises to an altitude where pressure is 27.5 kpa amd the temperature
inna [77]

Answer:

V₂ =  3227.46 L

Explanation:

Given data:

Initial volume of gas = 1000 L

Initial temperature = 50°C  (50 +273 = 323 K)

Initial pressure = 101.3 KPa

Final pressure = 27.5 KPa

Final temperature = 10°C  (10 +273 = 283 K)

Final volume = ?

Solution:

According to general gas equation:

P₁V₁/T₁ = P₂V₂/T₂

Formula:  

P₁V₁/T₁ = P₂V₂/T₂  

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

Now we will put the values in formula.

V₂ = P₁V₁ T₂/ T₁ P₂  

V₂ = 101.3 KPa × 1000 L × 283 K / 323 K × 27.5 KPa

V₂ = 28667900 KPa .L. K /   8882.5 K.KPa

V₂ =  3227.46 L

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