Based on reference table L, H2O is a stronger acid than

A sample of neon has a volume of 40.81 m3 at 23.5C. At what temperature, in Kelvins, would the gas occupy 50.00 cubic meters? As

mezya [45]

At $\fbox{\begin \\363 K \end{minispace}}$ temperature, a sample of neon gas will occupy $50.00 \text{ m}^{3}$ volume.

Further Explanation:

The given problem is based on the concept of Charles’ law. Charles’ law states that “at constant pressure and fixed mass the volume occupied an ideal gas is directly proportional to the Kelvin temperature.”

Mathematically the law can be expressed as,

$\fbox{ \begin \\ V \propto T \end{minispace}}$

Or,

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

Here, V is the volume of the gas, T is Kelvin temperature, and k is proportionality constant.

Given information:

The initial volume of neon gas is $40.81 \text{ m}^{3}$ .

The final volume of neon gas is $50.00 \text{ m}^{3}$ .

The initial temperature value is $23.5 \text{ } ^{\circ} \text{C}$ .

To calculate:

The final temperature

Given Condition:

The pressure is constant.

Mass of gas is fixed.

Solution:

Step 1: Modify the mathematical expression for Charles’ law for two different temperature and volume values as follows:

$\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}$

Here,

$V_{1}$ is the initial volume of the gas.

$V_{2}$ is the final volume of the gas.

$T_{1}$ is the initial temperature of the gas.

$T_{2}$ is the final temperature of the gas.

Step 2: Rearrange equation (2) for .

$\fbox {\begin \\T_{2}=\frac{(V_{2}) \times (T_{1})}{V_{1}}\\\end{minispace}}$ …… (2)

Step 3: Convert the given temperature from degree Celsius to Kelvin.

The conversion factor to convert degree Celsius to Kelvin is,

$T(\text{K}) = T(^{\circ}\text{C}) + 273.15$ …… (3)

Substitute $23.5\text{ }^{\circ} \text{C}$ for $T(^{\circ}\text{C})$ in equation (3) to convert temperature from degree Celsius to Kelvin.

$T(\text{K}) = 23.5 \text{ } ^{\circ} \text{C} + 273.15\\T(\text{K})= 296.65 \text{ K}$

Step 4: Substitute $40.81 \text{ m}^{3}$ for $V_{1}$ , $50.00 \text{ m}^{3}$ for $V_{2}$ and $296.65 \text{ K}$ for $T_{1}$ in equation (2) and calculate the value of $T_{2}$ .

$T_{2}=\frac{(50.00 \text{ m}^{3}) \times (296.65 \text{ K})}{40.81 \text{ m}^{3}}\\T_{2}=363.45 \text{ K}\\T_{2} \approx 363 \text{ K}$

Important note:

The temperature must be in Kelvin.

The condition of fixed mass and fixed pressure must be fulfilled in order to apply Charles’ law.

Learn More:

1. Gas laws brainly.com/question/1403211

2. Application of Charles’ law brainly.com/question/7434588

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: States of matter

Keywords: neon, volume, occupies, temperature, Kelvin, degree Celsius, Charle’s law, constant pressure, fixed mass, 40.81 m^3 , 50.00 m^3 , 23.5 degree C , celsius , 363 K , sates of matter, initial volume, final volume, initial temperature, final temperature, V1 , V2 , T1 , T2 .

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Valence Shell Electron Pair Repulsion theory is commonly known as VSEPR theory and it helps to predict the geometry of molecules.
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Molecular geometry us also octahedral with bond angles 90°.
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