Question

A balloon originally had a volume of 4.39 l at 44°c and a pressure of 729 torr. to what temperature must the balloon be cooled to reduce its volume to 3.78 l if the pressure is constant

Answer 1

Answer:

Explanation:

p*V/T = constant , but p = constant => V/T = constant

Note: T must be in Kelvin.

V1/T1 = V2/T2

T2 = (V2/V1)* T1

T2 = (3.78/4.39)* (44+273) = 272.95 ≈ 273 K = 0° C

The temperature must be 0° C

Answer 2

Considering the Charles's Law, the final temperature must be 273 K or 0 C.

Charles's Law consists of the relationship that exists between the volume and the temperature of a certain quantity of ideal gas, which is maintained at a constant pressure.

This law says that for a given sum of gas at constant pressure, as the temperature increases, the volume of the gas increases and as the temperature decreases, the volume of the gas decreases. That is, the volume and the temperature are directly proportional.

Mathematically, Charles's law says that when the amount of gas and pressure are kept constant, the quotient that exists between the volume and the temperature will always have the same value:

$\frac{V}{T} =k$

Studying two different states, an initial state 1 and a final state 2, it is satisfied:

$\frac{V1}{T1} =\frac{V2}{T2}$

In this case, you know:

• V1= 4.39 L
• T1= 44 C= 317 K (being O C= 273 K)
• V2= 3.78 L
• T2=?

Replacing in Charles's law:

$\frac{4.39 L}{317 K}=\frac{3.78 L}{T2}$

Solving:

$T2x\frac{4.39 L}{317 K}=3.78 L$

$T2=\frac{3.78 L}{\frac{4.39 L}{317 K}}$

T2= 273 K= 0 C

Finally, the final temperature must be 273 K or 0 C.

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