Respuesta :
Answer:
[tex]\theta_o = \frac{20 dm^3}{0.0741 C^{-1}}=269.91 C[/tex]
Step-by-step explanation:
Notation
[tex]V_i[/tex] represent the intercept for the linear model
[tex]\theta[/tex] represent the temperature in Celsius degrees
[tex]V[/tex] represent the volume for a given temperature on dm^3
For this case we have a linear model relating the volume V, with the temperature [tex]\theta[/tex] and with constant pressure.
So for this case the linear model would be:
[tex] V= 0.0741 \theta + V_i[/tex]
We have an initial condition given [tex] \theta=0 C, V= 20dm^3 [/tex]
If we use this condition in the linear model we can find the intercept:
[tex]20=0.0741*0+V_i [/tex]
So for this case the intercept is [tex]V_i =20[/tex]
So th linear model is given by:
[tex]V=0.0741 \theta +20[/tex]
We are interedtes on the absolute zero temperature, and that occurs when the volume V=0, and assuming this we can find the temperature [tex]\theta_o[/tex] where we have the absolute zero temperature. So we can set up the equation like this:
[tex]0=0.0741 \theta_o +20[/tex]
And solving for [tex]\theta_o[/tex] we got:
[tex]\theta_o = \frac{20 dm^3}{0.0741 C^{-1}}=269.91 C[/tex]
And that's our final answer.
