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Pepsi [2]
4 years ago
7

To what temperature was 32.0L of gas at 2°C be heated for it to occupy 1.00*10^2L at the same pressure?

Chemistry
1 answer:
jolli1 [7]4 years ago
8 0

Answer:

T₂ = 859.4 K

Explanation:

Given data:

Initial volume of gas = 32.0 L

Initial temperature = 2°C (2 + 273 = 275 k)

Final temperature = ?

Final volume = 1.00 ×10²L

Solution:

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

T₂ = V₂T₁ /V₁  

T₂ = 1.00 ×10²L × 275 K / 32.0 L

T₂ = 27500 L.K / 32.0 L

T₂ = 859.4 K

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