If the temperature of an ideal gas is raised from 100◦C to 200◦C, while the pressure remains constant, the volume 1. increases by a factor of 100. 2. None of these 3. doubles. 4. remains the same. 5. goes to 1 2 of the original volume.

Respuesta :

Answer:

3. doubles

Explanation:

for an ideal gas behavior, the relationship between volume and temperature is given by Charles law

Charles law states that the volume of a given mass of gas is directly proportional to its temperature provided that pressure remains constant. Mathematically, this is represented as

V ∝ T

V=KT

K = V/T

where V is the volume of the gas

T is the Temperature

k represents the constant of proportionality

For initial and final conditions of a gas,

[tex]\frac{V_{1} }{T_{1} }[/tex] = [tex]\frac{V_{2} }{T_{2} }[/tex]

where 1 and 2 represent initial and final conditions respectively

therefore, T₁ = 100 and T₂ = 200

[tex]\frac{V_{1} }{100}[/tex] = [tex]\frac{V_{2} }{200}[/tex]

200 × V₁ = 100 × V₂

divide both sides by 100

2V₁ = V₂

final volume,V₂ = 2V₁

there the volume doubles

The increase in temperature results in the volume of the gas to be doubled. Hence, option 3 is correct.

The ideal gas relationship with volume and temperature at constant pressure has been given by Charles law. The relation has been expressed as:

[tex]\dfrac{V_1}{T_1} =\dfrac{V_2}{T_2}[/tex]

Where the initial volume of gas, [tex]V_1[/tex]

The final volume of gas, [tex]V_2[/tex]

The initial temperature of gas, [tex]T_1=100^\circ \rm C[/tex]

The final temperature of gas, [tex]T_2=200^\circ \rm C[/tex]

Substituting the values for relation between volume has been given as:

[tex]\dfrac{V_1}{100} =\dfrac{V_2}{200}\\V_2=\dfrac{V_1}{100} \;\times\;200\\V_2=2\;V_1[/tex]

The final volume gas has been double the initial volume. Thus, with increase in temperature the volume of the gas as been doubled. Hence, option 3 is correct.

For more information about ideal gas, refer to the link:

https://brainly.com/question/8711877