A tank contains helium for filling balloons. It holds 6.75 L of Ne at 24.8 atm
and 27.5 °C. What would be the volume of this helium at 2.13 atm and 10.6 °C?
(P1V1/T1) = (P2V2/T2)
To find the volume of helium at a different pressure and temperature, we can use the Combined Gas Law equation:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
Where:
P1 = Initial pressure (24.8 atm)
V1 = Initial volume (6.75 L)
T1 = Initial temperature (27.5 °C + 273.15 = 300.65 K)
P2 = Final pressure (2.13 atm)
V2 = Final volume (to be determined)
T2 = Final temperature (10.6 °C + 273.15 = 283.75 K)
Plug these values into the equation:
(24.8 atm * 6.75 L) / (300.65 K) = (2.13 atm * V2) / (283.75 K)
Now solve for V2:
V2 = (24.8 atm * 6.75 L * 283.75 K) / (300.65 K * 2.13 atm)
V2 = (5099.5 atm*L*K) / (641.1345 atm*K)
V2 ≈ 7.95 L
Therefore, the volume of helium at 2.13 atm and 10.6 °C would be approximately 7.95 L.