How is wind deflation different from wind abrasion??
1 answer:
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Answer:
1090 mmHg
Explanation:
We know that with gases we must use a Kelvin temperatures, so let’s try a plot of pressure against the Kelvin temperature.
We can create a table as follows
<u>t/°C</u> <u>T/K</u> <u>p/mmHg</u>
10 283 726
20 293 750
40 313 800
70 343 880
100 373 960
150 423 ???
I plotted the data and got the graph in the figure below.
It appears that pressure is a linear function of the Kelvin temperature.
y = mx + b
where x is the slope and b is the y-intercept.
===============
<em>Calculate the slope </em>
I will use the points (275, 700) and (380, 975).
Slope = Δy/Δx = (y₂ - y₁)/(x₂ -x₁) = (975 -700)/(380 – 275) = 275/105 = 2.619
So,
y = 2.619x + b
===============
<em>Calculate the intercept </em>
When x = 275, y = 700.
700 = 2.619 × 275 + b
700 = 720 + b Subtract 720 from each side and transpose.
b = -20
So, the equation of the graph is
y = 2.619x -20
===============
<em>Calculate the pressure</em> at 423 K (150°C)
y = 2.619 × 423 - 20
y = 1110 - 20
y = 1090
At 150 °C, the pressure 1090 mmHg.
The point is approximately at the position of the black dot in the graph.
