Question

A cylinder 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

Answer 1

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: 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, 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.