These data are from a constant-volume gas thermometer experiment. The volume of the gas was kept constant, while the temperature was changed. The resulting pressure was measured. Plot the data on a pressure versus temperature diagram. Based on these data, estimate the value of absolute zero in Celsius. $$\begin{array}{cc} \hline T\left(^{\circ} \mathrm{C}\right) & P(\mathrm{atm}) \\\ 0 & 1.00 \\\ 20 & 1.07 \\\ 100 & 1.37 \\\ -33 & 0.88 \\\ -196 & 0.28 \\\ \hline \end{array}$$

Short Answer

Expert verified
Answer: The approximate value of absolute zero based on the plotted data and the calculated linear function is -222.22°C.

Step by step solution

01

Plot the given data

Create a graph, with the x-axis representing the temperature in Celsius and the y-axis representing the pressure in atmospheres. Plot each of the given points on the graph: - \((0,1.00)\) - \((20,1.07)\) - \((100,1.37)\) - \((-33,0.88)\) - \((-196,0.28)\)
02

Draw the best-fit line

Draw a straight line that passes through the points, or as close to each point as possible. This line should represent the relationship between the pressure and temperature.
03

Find the linear function representing the best-fit line

Calculate the slope (m) and the y-intercept (b) of the best-fit line. Let's say the calculated values are \(m=0.0045\) and \(b=1.00\) (Note: these values may vary depending on the drawn line). This linear function will be represented as: $$P = 0.0045T + 1.00$$ where P is the pressure in atmospheres and T is the temperature in Celsius.
04

Estimate the value of absolute zero

The absolute zero is the temperature at which the pressure becomes zero. To find the value of absolute zero, set P equal to 0 in the linear function and solve for T: $$0 = 0.0045T + 1.00$$ Solving for T: $$T = \frac{-1.00}{0.0045} = -222.22 ^{\circ}\mathrm{C}$$ The estimate of absolute zero is approximately \(-222.22^{\circ}\)C. However, it's important to remember that this value is dependent upon your drawn best-fit line, and the actual value of absolute zero is \(-273.15^{\circ}\)C.

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