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What change in volume results if 50.0 mL of gas is cooled from 48.0 °C to
3°C?

1 Answer

2 votes

Answer:

-2.6 mL.

Step-by-step explanation:

To solve this question, we need to use the formula:

V1/T1 = V2/T2

where V1 and T1 are the initial volume and temperature of the gas, and V2 and T2 are the final volume and temperature of the gas. We also need to convert the temperatures from degrees Celsius to kelvins by adding 273.15. Plugging in the given values, we get:

50.0 mL / (48.0 + 273.15) K = V2 / (3 + 273.15) K

Solving for V2, we get:

V2 = 50.0 mL x (3 + 273.15) K / (48.0 + 273.15) K V2 = 47.4 mL

Therefore, the change in volume is:

ΔV = V2 - V1 ΔV = 47.4 mL - 50.0 mL ΔV = -2.6 mL

The negative sign indicates that the volume decreases when the gas is cooled.

The answer is -2.6 mL.

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