Question

A gas sample has an initial volume of 63.2 mL, an initial temperature of 42.0 ?C, and an initial pressure of 751 mmHg. The volume is decreased to 47.6 mL and the temperature is increased to 77.0 ?C.

Answer 1

Answer:

Lets Write Down the Given Initial Conditions.

$P_1 = 751mmHg$           $P_2 = --$

$V_1 = 63.2ml$                $V_2 = 47.6ml$

$T_1 = 42C$                     $T_2 = 77$

In Order to Solve for the Unknown: $P_2$

we must use the Ideal Gas Law to Solve for the Second Unknown pressure:

$\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}$

Then Rearrange this equation in a form where P2 can be solved from:

$P_2 = \frac{P_1V_1T_2}{T_1V_2}$

Then Insert the Values from above to solve:

$P_2 = \frac{(751 mmHg)(63.2ml)(77C)}{(47.6ml)(42C)}$

The Answer is : 1830 mmHg considering sig figs