A technician is monitoring the pressure of a gas inside a container during an experiment. … — SAT

A technician is monitoring the pressure of a gas inside a container during an experiment. The pressure $P$ , in kilopascals ( $\text{kPa}$ ), and the time elapsed $t$ , in minutes, are related by a linear equation. At $t = 4$ minutes, the pressure is $112\text{ kPa}$ . At $t = 12$ minutes, the pressure is $136\text{ kPa}$ . If the pressure continues to increase at this constant rate, what will the pressure be, in $\text{kPa}$ , at $t = 15$ minutes?

Answer: 145 kPa

Answer

145

The relationship between pressure

P

and time

t

is linear, which can be modeled by the equation

P = mt + b

, where

m

is the rate of change (slope) and

b

is the pressure at

t = 0

. Using the points

(4, 112)

and

(12, 136)

, the slope is calculated as

m = \frac{136 - 112}{12 - 4} = \frac{24}{8} = 3

. Substituting the point

(4, 112)

and

m = 3

into the equation

P = mt + b

yields

112 = 3(4) + b

, which simplifies to

112 = 12 + b

, so

b = 100

. The linear equation is

P = 3t + 100

. Substituting

t = 15

into this equation gives

P = 3(15) + 100 = 45 + 100 = 145

Step-by-Step Solution

Calculate the rate of change (slope,

m

) using the two given coordinate points

(4, 112)

and

(12, 136)

m = \frac{136 - 112}{12 - 4} = \frac{24}{8} = 3

To find the constant rate at which the pressure is increasing per minute.

Set up the linear equation using the point-slope form

P - P_1 = m(t - t_1)

with the point

(4, 112)

P - 112 = 3(t - 4) \Rightarrow P = 3t + 100

To establish the linear relationship between pressure and time.

Substitute

t = 15

into the linear equation to find the pressure at

15

minutes.

P = 3(15) + 100 = 145

To determine the pressure at the requested time of

15

minutes.

Key Concept

Finding and applying a linear equation in two variables from two points.

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