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saw5 [17]
3 years ago
9

sample of carbon monoxide gas occupies 3.20 L at 125 °C. At what temperature will the gas occupy a volume of 1.54 L if the press

ure remains constant?
Chemistry
2 answers:
Anna [14]3 years ago
3 0

Answer:

-81.5 degrees C or 191.5 K

Explanation:

We want to use Charles' gas law: V/T = V/T

Our initial volume is 3.20 L, and our initial temperature is 125 degrees C, or 125 + 273 = 398 degrees Kelvin.

Our new Volume is 1.54 L, but we don't know what the temperature is. So, we use the equation:

3.20 L / 398 K = 1.54 L / T ⇒ Solving for T, we get: T = 191.5 K

If we want this in degrees Celsius, we subtract 273: 191.5 - 273 = -81.5 degrees C

Sophie [7]3 years ago
3 0

Answer:

- 81.5°C

Explanation:

Data obtained from the question include:

V1 (initial volume) = 3.20 L

T1 (initial temperature) = 125 °C = 125 + 273 = 398K

V2 (final volume) = 1.54 L

T2 (final temperature) =.?

At constant pressure indicates that the gas is obeying Charles' law. Using the Charles' law equation V1/T1 = V2/T2 , the final temperature of the gas can be obtained as illustrated below:

V1/T1 = V2/T2

3.2/398 = 1.54/T2

Cross multiply to express in linear form

3.2 x T2 = 398 x 1.54

Divide both side by 3.2

T2 = (398 x 1.54)/ 3.2

T2 = 191.5K

Now let us convert 191.5K to celsius temperature. This is illustrated below

°C = K - 273

°C = 191.5 - 273

°C = - 81.5°C

Therefore the gas will occupy 1.54L at a temperature of - 81.5°C

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CK-12 Boyle and Charles's Laws if Mrs. Pa pe prepares 12.8 L of laughing gas at 100.0 k Pa and -108 °C and then she force s the
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Answer:

The answer to your question is   P2 = 2676.6 kPa

Explanation:

Data

Volume 1 = V1 = 12.8 L                        Volume 2 = V2 = 855 ml

Temperature 1 = T1 = -108°C               Temperature 2 = 22°C

Pressure 1 = P1 = 100 kPa                    Pressure 2 = P2 =  ?

Process

- To solve this problem use the Combined gas law.

                     P1V1/T1 = P2V2/T2

-Solve for P2

                     P2 = P1V1T2 / T1V2

- Convert temperature to °K

T1 = -108 + 273 = 165°K

T2 = 22 + 273 = 295°K

- Convert volume 2 to liters

                       1000 ml -------------------- 1 l

                         855 ml --------------------  x

                         x = (855 x 1) / 1000

                         x = 0.855 l

-Substitution

                    P2 = (12.8 x 100 x 295) / (165 x 0.855)

-Simplification

                    P2 = 377600 / 141.075

-Result

                   P2 = 2676.6 kPa

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