An ideal gas is confined to a container with adjustable volume. The pressure and mole number are constant. By what factor will volume change if absolute temperature doubles

P V = n R T

so
V/T = n R/P
right is constant so left is constant
if the denominator doubles, the numerator must double.

According to the Ideal Gas Law, the relationship between pressure (P), volume (V), mole number (n), and absolute temperature (T) for an ideal gas is given by the equation:

PV = nRT

Where R is the gas constant.

In this case, we are told that the pressure and mole number are constant. Therefore, we can rewrite the equation as:

V = (nR/P) * T

We are asked to determine the change in volume if the absolute temperature doubles. Let's call the initial temperature T1 and the final temperature T2. We can write the equation for the initial and final volumes as:

V1 = (nR/P) * T1
V2 = (nR/P) * T2

We can find the factor by which the volume changes by dividing V2 by V1:

(V2/V1) = [(nR/P) * T2] / [(nR/P) * T1]

Simplifying the equation, we find:

(V2/V1) = T2/T1

Since we are doubling the absolute temperature (i.e., T2 = 2T1), we can substitute this into the equation:

(V2/V1) = (2T1) / T1

Canceling out T1, we find:

(V2/V1) = 2

Therefore, the volume will double if the absolute temperature doubles.

To determine the factor by which volume changes when the absolute temperature doubles for an ideal gas with constant pressure and mole number, we can use the ideal gas law.

The ideal gas law is given by the equation: PV = nRT

Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant
T = absolute temperature

In this case, we are given that the pressure (P) and the number of moles (n) are constant. Therefore, we can rewrite the equation as:

V₁T₁ = V₂T₂

Where:
V₁ = initial volume
T₁ = initial absolute temperature
V₂ = final volume
T₂ = final absolute temperature

Since we are doubling the absolute temperature (T₂ = 2T₁), we can substitute this into the equation:

V₁T₁ = V₂(2T₁)

Now, we can solve for the final volume (V₂):

V₂ = V₁T₁ / (2T₁)
V₂ = V₁ / 2

Therefore, the factor by which the volume changes when the absolute temperature doubles is 1/2.