Chemistry, 23.08.2019 21:40 sherry59Sherry59

Acylinder with a movable piston contains a sample of gas having a volume of 6.0 liters at 293 k and 1.0 atmosphere. what is the volume of the sample after the gas is heated to 303 k, while the pressure is held at 1.0 atmosphere?
(1) 9.0 l (3) 5.8 l
(2) 6.2 l (4) 4.0 l

Answers: 2

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Answer from: drakkeo

The volume of sample when the gas is heated to 303 K is $\boxed{{\text{(2)6}}{\text{.2 L}}}$

Further explanation:

Charles’s law:

Charles’s work showed that at constant pressure, the volume-temperature relationship for a fixed amount of gas is linear. In other words, Charles’s law can be stated that at constant pressure, the volume occupied by a fixed amount of a gas is directly proportional to its absolute temperature (Kelvin). This relationship is known as Charles’s law.

The mathematical representation of Charles’s law is,

${\mathbf{V}}\propto{\mathbf{T}}$ [P and n are constant]

Where,

⦁V is volume occupied by the fixed quantity of gas.

⦁T is the temperature of a gas.

⦁P is the pressure of a gas.

⦁n denotes the number of moles of gas.

The relationship can also be expressed as,

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

Or it can also be expressed as follows:

$\frac{{{{\text{V}}_{\text{1}}}}}{{{{\text{T}}_{\text{1}}}}} = \frac{{{{\text{V}}_{\text{2}}}}}{{{{\text{T}}_{\text{2}}}}}$ …… (1)

Here,

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

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

${{\text{T}}_1}$ is the initial temperature of the gas.

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

Rearrange equation (1) to calculate ${{\text{V}}_2}$ .

${{\text{V}}_{\text{2}}} = \frac{{{{\text{V}}_{\text{1}}}{{\text{T}}_{\text{2}}}}}{{{{\text{T}}_{\text{1}}}}}$ …… (2)

The value of ${{\text{V}}_1}$ is 6 L.

The value of ${{\text{T}}_1}$ is 293 K.

The value of ${{\text{T}}_2}$ is 303 K.

Substitute these values in equation (2).

$\begin{aligned}{{\text{V}}_{\text{2}}}&=\frac{{\left({{\text{6 L}}}\right)\left( {{\text{303 K}}}\right)}}{{\left( {{\text{293 K}}}\right)}}\\&={\text{6}}{\text{.20477 L}}\\&\approx{\mathbf{6}}{\mathbf{.2 L}}\\\end{aligned}$

So option (2) is the correct answer.

Learn more:

1. Law of conservation of matter states: link

2. Calculation of volume of gas: link

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Ideal gas equation

Keywords: Charles’s law, volume, temperature, pressure, volume temperature relationship, absolute temperature, constant pressure, relationship, V directly proportional to T, ideal gas, ideal gas equation number of moles, moles, P, n, V, T, volume of gas, 6.2 L, 303 K, 293 K, 6 L.

Answer from: Quest

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Answer from: Quest

We first assume that this gas is an ideal gas where it follows the ideal gas equation. the said equation is expressed as: pv = nrt. from this equation, we can predict the changes in the pressure, volume and temperature. if the volume and the temperature of this gas is doubled, then the pressure still stays the same.

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Acylinder with a movable piston contains a sample of gas having a volume of 6.0 liters at 293 k and...