Question

1) The data in the table below represents the pressure of the gas as the temperature changes. Plot a graph of the data, using the space below. Draw a trend line and calculate its slope. How are the variables related? What will he pressure of the gas be at 0C?

the graph looks like this
Temperature Pressure
10 723
25 763
40 800
55 845
80 900

Answer 1

The pressure of the gas 0°C is about 699.

Explanation:

The given table of values is

Temperature Pressure

10 723

25 763

40 800

55 845

80 900

The general form of a trend line is

`y=a+bx`

where, b is slope and a is y-intercept.

`b=\frac{\sum_(i=1)^nx_iy_i-n\overline{x}\overline{y}}{\sum_(i=1)^nx_i^2-n\overline{x}^2}`

`a=\overline{y}-b\overline{x}`

Using graphing calculator, we get

`a=699.006826`

`b=2.552218`

The equation of line of best bit is

`y = 2.552218x + 699.006826`

where, y is pressure at x°C.

We need to find the pressure of the gas at 0°C.

Substitute x=0 in the above equation.

`y = 2.552218(0) + 699.006826`

`y =699.006826`

`y \approx 699`

Therefore the pressure of the gas 0°C is about 699.

Answer 2

The graph of temperature versus pressure is a straigh line rising to the right. It Using the value above, the equation is P = 2.5522T + 699. The temperature is directly proportional to pressure. The pressure of the gas 0C using the equation,

P = 2.5522T + 699
P = 2.5522(0) + 699
P = 699