If the pressure is 0.975 atm, what is the temperature?

You are researching the gases that keep lily pads afloat. The cell vacuoles that contain the oxygen that keeps the plant afloat are approximately 17.7 μL. They can expand and contract with changes in temperature and pressure, but that is the optimal volume at 25 °C and standard pressure. If the volume drops below 14.2 μL the leaves are submerged. You observe the lily pads in the pond are below water level, so the volume is down. If the pressure is 0.975 atm, what is the temperature?

If the pressure is 0.975 atm, what is the temperature?
Use the combined gas law to find the final temperature, T2. The combined gas law is stated as: P1V1T2=P2V2T1. Here are your given data:

P1=std pressure, w/c is 1 atm

V1=17.7uL

T2= x

P2=0.975 atm

V2=14.2uL

T1=25 C or 298K; remember to always convert to K before proceding to your computations.



Rearranging the above formula gives you:

T2 = (P2*V2*T1)/(P1*V1)

= 233.1K or -39.9 C.
Reply:Combined gas law.



P1V1T2 = P2V2T1

P1 = 1atm. V1 = 17.7 μL. T2 = ?

P2 = 0.975atm. V2 = 14.2 μL. T1 = 298K



T2 = (0.975 x 14.2 x 298) ÷ (1 x 17.7)

T2 = 4125.81 ÷ 17.7

T2 = 233.1K = -39.9°C

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