# 3.20 cm3 air bubble forms in a deep lake at a depth where the temperature is 8°C at a total pressure...?

A 3.20 cm3 air bubble forms in a deep lake at a depth where the temperature is 8°C at a total pressure of 2.45 atm. The bubble rises to a depth where the temperature and pressure are 19°C and 1.12 atm, respectively. Assuming the amount of air in the bubble has not changed, calculate its new volume.

We use the equation :

p1V1 / T1 = p2 V2 / T2

V1 = 3.20 cm^3 = 3.20 mL = 0.0032 L

T1 = 273 + 8 = 281 K

p1 = 2.45 atm

V2 = ?

T2 = 273 + 19 = 292 K

p2 = 1.12 atm

2.45 x 0.0032 / 281 = 1.12 x V2 / 292

V2 = 0.00727 L = 7.27 mL = 7.27 cm^3

PV=nRT. Yes, it's that simple. Heck, it's even simpler if you think about the constants in the equation.

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**Answer:**We use the equation :

p1V1 / T1 = p2 V2 / T2

V1 = 3.20 cm^3 = 3.20 mL = 0.0032 L

T1 = 273 + 8 = 281 K

p1 = 2.45 atm

V2 = ?

T2 = 273 + 19 = 292 K

p2 = 1.12 atm

2.45 x 0.0032 / 281 = 1.12 x V2 / 292

V2 = 0.00727 L = 7.27 mL = 7.27 cm^3

PV=nRT. Yes, it's that simple. Heck, it's even simpler if you think about the constants in the equation.

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