A balloon, whose volume at 23°C is 535 mL, is heated to 46°C. Assuming the pressure and amount of gas remain constant, what is the volume of the balloon at 46°C?

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11-04-2014
Chemistry

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A balloon, whose volume at 23°C is 535 mL, is heated to 46°C. Assuming the pressure and amount of gas remain constant, what is the volume of the balloon at 46°C?

Sagot :

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KDR KDR · Answer 1 · 2014-04-12T15:44:11-04:00

This problem requires a certain equation. That equation is V1/T1=V2/T2, where V1 is your initial volume (535 mL in this case), T1 is your initial temperature in Kelvin(23 degrees C = 296 K), V2 is your final volume (unknown), and T2 is your final temperature (46 degrees C = 319 K). By plugging in these values, the equation looks like this: 535/296=V2/319. Now multiply both sides of the equation by 319, and your final answer is V2= 576.6 mL

avantikar avantikar · Answer 2 · 2020-02-26T06:06:50-05:00

The volume of balloon at [tex]46\text{ }^{\circ}\text{C}[/tex] is [tex]\boxed{576.5\text{ mL}}[/tex].

Further explanation:

Charles’s law:

This law describes volume-temperature relationship of gases at constant amount of gas and pressure. According to this law, volume occupied by a fixed amount of a gas is directly proportional to its absolute temperature, provided pressure and amount of gas is kept constant.

Mathematical expression of Charles’s law:

[tex]\text{V}\propto\text{T}[/tex]

Here,

V is volume occupied by the fixed quantity of gas.

T is temperature of gas.

The relationship can also be expressed as follows:

[tex]\dfrac{\text{V}}{\text{T}}=\text{constant}[/tex] [P and n are constant]

Or it can also be expressed as follows:

[tex]\dfrac{\text{V}_1}{\text{T}_1}=\dfrac{\text{V}_2}{\text{T}_2}[/tex] ...... (1)

Here,

[tex]\text{V}_1[/tex] is initial volume of gas.

[tex]\text{V}_2[/tex] is final volume of gas.

[tex]\text{T}_1[/tex]is initial temperature of gas.

[tex]\text{T}_2[/tex] is final temperature of gas.

Rearrange equation (1) for [tex]\text{V}_2[/tex] .

[tex]\text{V}_2=\dfrac{\text{V}_1\text{T}_2}{\text{T}_1}[/tex] ...... (2)

The value of [tex]\text{T}_1[/tex] can be calculated as follows:

[tex]\begin{aligned}\text{T}_1&=\left(23+273.15\right)\text{K}\\&=296.15\text{ K}\end{aligned}[/tex]

The value of [tex]\text{T}_2[/tex] can be calculated as follows:

[tex]\begin{aligned}\text{T}_2&=\left(46+273.15\right)\text{K}\\&=319.15\text{ K}\end{aligned}[/tex]

Substitute 296.15 K for [tex]\text{T}_1[/tex], 319.15 K for [tex]\text{T}_2[/tex] and 535 mL for [tex]\text{V}_1[/tex] in equation (2).

[tex]\begin{aligned}\text{V}_2&=\dfrac{\left(535\text{ mL}\right\left(319.15\text{ K}\right)}{\left(296.15\text{ K}\right)}\\&=576.5\text{ mL}\end{aligned}[/tex]

Therefore final volume of balloon comes out to be 576.5 mL.

Learn more:

1. Law of conservation of matter states: https://brainly.com/question/2190120

2. Calculation of volume of gas: https://brainly.com/question/3636135

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Ideal gas equation

Keywords: Charles’s law, volume, temperature, pressure, volume-temperature relationship, absolute temperature, 576.5 mL, 296.15 K, 319.15 K, 535 mL, T2, T1, V1, V2, 296.15 K, 319.15 K.

Sagot :

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