Boyle's Law-effect of pressure at constant temperature imagine that you performed this experiment as described in your manual. In doing so, your first pressure reading was 660 mmHg, and your second presure reading was 780 mmig Your first volume was 474 mL What would the percent error be if your measured second volume was 38.2 mL ? 4.76 X incorrect, One attempt remaining

Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

[tex]P\propto \frac{1}{V}[/tex]

or,

[tex]P_1V_1=P_2V_2[/tex]

where,

[tex]P_1[/tex] = first pressure = 660 mmHg

[tex]P_2[/tex] = second pressure = 780 mmHg

[tex]V_1[/tex] = first volume = 47.4 mL

Note : (The value of first volume might be 47.4 mL because the difference between the value of first and second volume can not be more.)

[tex]V_2[/tex] = second volume = ?

Now put all the given values in the above equation, we get:

[tex]660mmHg\times 47.4mL=780mmHg\times V_2[/tex]

[tex]V_2=40.1mL[/tex]

Thus, the value of second volume is, 40.1 mL

Now we have to calculate the percent error.

To calculate the percentage error, we use the equation:

[tex]\%\text{ error}=\frac{|\text{Experimental value - Accepted value}|}{\text{Accepted value}}\times 100[/tex]

We are given:

Experimental value of volume = 38.2 mL

Accepted value of volume = 40.1 mL

Putting values in above equation, we get:

[tex]\%\text{ error}=\frac{|40.1-38.2|}{40.1}\times 100\\\\\%\text{ error}=4.74\%[/tex]

Hence, the percentage error will be, 4.74%