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antoniya [11.8K]
3 years ago
7

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 t

he volume of the balloon at 46°C?    
Chemistry
2 answers:
NikAS [45]3 years ago
5 0
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
Gelneren [198K]3 years ago
5 0

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

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:

\text{V}\propto\text{T}

Here,

V is volume occupied by the fixed quantity of gas.

T is temperature of gas.

The relationship can also be expressed as follows:

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

Or it can also be expressed as follows:

\dfrac{\text{V}_1}{\text{T}_1}=\dfrac{\text{V}_2}{\text{T}_2}                                                                           ...... (1)

Here,

\text{V}_1 is initial volume of gas.

\text{V}_2 is final volume of gas.

\text{T}_1is initial temperature of gas.

\text{T}_2 is final temperature of gas.

Rearrange equation (1) for \text{V}_2 .

\text{V}_2=\dfrac{\text{V}_1\text{T}_2}{\text{T}_1}                                                                           ...... (2)

The value of \text{T}_1 can be calculated as follows:

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

The value of \text{T}_2 can be calculated as follows:

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

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

\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}

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

Learn more:

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

2. Calculation of volume of gas: 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.

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a) CO₂: <em>21,9%; </em>CO: <em>7,0%; </em>CH₄: <em>18,2%; </em>H₂: <em>0,8%; </em>N₂: <em>52,1%</em>

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Explanation:

a) It is posible to obtain the composition of the gas mixture in weight% using molecular mass of each compound, thus:

12% CO₂×\frac{44,01g}{1mol} = <em>528,1 g</em>

6% CO×\frac{28,01g}{1mol} = <em>168,1 g</em>

27,3% CH₄×\frac{16,05g}{1mol} = <em>438,2 g</em>

9,9% H₂×\frac{2,02g}{1mol} = <em>20,0 g</em>

44,8% N₂×\frac{28g}{1mol} = <em>1254,4 g</em>

The total mass of the gas mixture is:

528,1g + 168,1g + 438,2g + 20,0g + 1254,4g = <em>2408,8 g</em>

Thus composition of the gas mixture in weight% is:

CO₂: \frac{528,1g}{2408,8g}×100 = <em>21,9%</em>

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CH₄: \frac{438,2g}{2408,8g}×100 = <em>18,2%</em>

H₂: \frac{20,0g}{2408,8g}×100 = <em>0,8%</em>

N₂: \frac{1254,4g}{2408,8g}×100 = <em>52,1%</em>

b) The average molecular weight of the gas mixture is determined with mole % composition, thus:

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I hope it helps!

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