Answer to Question #176072 in Physical Chemistry for JAE

Question

A helium filled balloon at sea level has a volume of 2.1 L at 0.998 atm and 36℃.  If it is released and rises to an elevation at which the pressure is 0.900 atm and the temperature is 28℃, what will be the new volume of the balloon?

An unopened, cold 2.00 L bottle of soda contains 46.0 mL of gas confined at a pressure of 1.30 atm at the temperature of 5.0℃.  If the bottle is dropped into a lake and sinks to a depth at which the pressure is 1.52 atm and the temperature is 2.09℃, what will be the volume of gas in the bottle?

A sample of air in a syringe exerts a pressure on 1.02 atm at a temperature of 22.0℃  The syringe is placed in a boiling water bath at 100.0℃.  The pressure of the air is increased to 1.23 atm by pushing the plunger in, which reduced the volume to 0.224 mL  What was the original volume of the air?

I NEED THESE BY 10PM PACIFIC TIME

Answer 1

New balloon volume V2

$\frac{P1V1}{T1}= \frac{P2V2}{T2}$

$V2= \frac{P1V1T2}{T1P2}$

V2= $\frac{ 0.998×2.1×(28+273)}{(36+273)×0.900}=2.27L$

New volume of gass

$\frac{P1V1}{T1}= \frac{P2V2}{T2}$

$V2= \frac{P1V1T2}{T1P2}$

V2= $\frac{ 1.30×46.0×(2.09+273)}{(5.0+273)×1.52}=38.9mL$

The original volume

$\frac{P1V1}{T1}= \frac{P2V2}{T2}$

$V1= \frac{P2V2T1}{T2P1}$

V1 = $\frac{ 1.23×0.224×(22.0+273)}{(100+273)×1.02}=0.214mL$