Mslope 3263 Byintercept 2415 Correlation 09951 RMSE 3593 1

In order to determine an experimental fixed amount of air ec re values for the trapped gas, she obtained the following data: value for absolute zero, a student unt of air into a beaker of water. By varying the temperature submerged a flask containing a the corresponding and measuring pressure Pressure (atm) Temperature (C) 0.966 0.911 0.857 0.802 0.747 80.0 60.0 40.0 20.0 0.0 these data into Logger Pro (Pressure in X and Temperature in Y). Perform a linear regression, L.e., the equation of the straight line that best fits the data (see Graphing exerise, page 24) Record the slope and intercept of the line of best fit: 321.30am b(intercept):--2415 m (slope): Yan b (intercept): = ecord the \"goodness of fit\" parameters: Corelaton9951 RMSE3.63C Correlation rom this regression analysis, what temperature corresponds to zero pressure (i.e., what temperature in °C corresponds to absolute zero). (ans: -273 C) Using the equation of the line, solve for the temperature of the trapped gas that corresponds to a pressure of 0.488 atm (ans: -94.8°C Using the equation of the line, solve for the pressure of the trapped gas when the temperature is lowered t 40.0°C (ans: 0.638 atn

Solution

1.

Using technology, we get              
              
slope =    365.627047          
intercept =    -273.1961285          
              
Thus, the regression line is              
              
y^ =    365.627047   x   -   273.1961285

Thus, if x = 0,

y^ = -273.1961285 degrees C [ANSWER]

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2.

The regression line is              
              
y^ =    365.627047   x   -   273.1961285
              
Thus, if x =    0.488 atm      
              
Then              
              
y^ =    -94.77012954 degrees C [ANSWER]

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3.

The regression line is              
              
y^ =    365.627047   x   -   273.1961285

-40 =    365.627047   x   -   273.1961285

Solving for x,

x = 0.637797806 atm [ANSWER]