What is the volume of the balloon at this depth?

Question

A scuba diver takes a 2.3 L balloon from the surface, where the pressure is 1.1 atm and the temperature is 31 ∘C, to a depth of 25 m, where the pressure is 3.5 atm and the temperature is 14 ∘C.

Roger the Mole · Accepted Answer

(2.3 L) x (1.1 atm / 3.5 atm) x (14 + 273) K / (31 + 273) K = 0.68 L

Jim · Answer

PV=nRT <<<memorize!!
n is moles, R is gas constant (use correct one!), T is temp in K (C+273.15)
Since you have a before and after with the same n &R simplifies to
PV/T initial = nR = PV/T final
Now sub in your numbers and solve
SI Units R values:
8.31446261815324 J⋅K−1⋅mol−1
8.31446261815324 m3⋅Pa⋅K−1⋅mol−1
8.31446261815324 kg⋅m2·K−1⋅mol−1s−2
8.31446261815324×103 L⋅Pa⋅K−1⋅mol−1
8.31446261815324×10−2 L⋅bar⋅K−1⋅mol−1
0.082057 L atm mol-1K-1
62.36L·torr/mol·K = L·mmHg/mol·K

nyphdinmd · Answer

pV = nRT  -->  note n*R = constant so

Let V0, p0, and T0 be volume, pressure and temperature at surface (respectively) and

V1, p1, T1 be volume, pressure and temperature at depth (respectively)

Using ideal gas equation of state  nR = p0*V0/T0

then using it again for the balloon at depth

p1*V1 = nRT1 = {p0*V0/T0} T1  -->  solve for V1

V1 = {p0*V0/(p1*T0}}*T1

Plug in values and you're finished

What is the volume of the balloon at this depth?

3 Answers