Question

A certain sample of a gas has a volume of 20.00 dm3 at 0 °C and 1.000 atm. A plot of the experimental data of its volume against the Celsius temperature, θ, at constant p, gives a straight line of slope 0.0741 dm3 °C−1. From these data alone (without making use of the perfect gas law), determine the absolute zero of temperature in degrees Celsius.

Answer 1

Answer:

$\theta_o = \frac{20 dm^3}{0.0741 C^{-1}}=269.91 C$

Step-by-step explanation:

Notation

$V_i$ represent the intercept for the linear model

$\theta$ represent the temperature in Celsius degrees

$V$ 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 $\theta$ and with constant pressure.

So for this case the linear model would be:

$V= 0.0741 \theta + V_i$

We have an initial condition given $\theta=0 C, V= 20dm^3$

If we use this condition in the linear model we can find the intercept:

$20=0.0741*0+V_i$

So for this case the intercept is $V_i =20$

So th linear model is given by:

$V=0.0741 \theta +20$

We are interedtes on the absolute zero temperature, and that occurs when the volume V=0, and assuming this we can find the temperature $\theta_o$ where we have the absolute zero temperature. So we can set up the equation like this:

$0=0.0741 \theta_o +20$

And solving for $\theta_o$ we got:

$\theta_o = \frac{20 dm^3}{0.0741 C^{-1}}=269.91 C$

And that's our final answer.