A compressed air tank carried by scuba divers has a volume of 9.5 L and a pressure of 140 atm at 20∘C.
If the gas was instead in a cylinder with a floating, massless, frictionless piston, what would the volume of the gas be (in liters) at STP?
Express the volume in liters to two significant digits.
(P1V1/T1) = (P2V2/T2)
1.2x10^3 L
To find the volume of the gas at STP (Standard Temperature and Pressure), we need to apply the ideal gas law.
The ideal gas law is given by the equation: PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles of gas
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (in Kelvin)
First, we need to convert the pressure and temperature to the units used in the ideal gas law.
Given:
Pressure (P) = 140 atm
Temperature (T) = 20°C = 293.15 K (20 + 273.15)
Since the volume is already given in liters, we do not need to make any conversions for it.
Now, let's solve for n, the number of moles of gas using the ideal gas law:
n = (PV) / (RT)
Substituting the given values:
n = (140 atm * 9.5 L) / (0.0821 L·atm/(mol·K) * 293.15 K)
Simplifying the equation:
n ≈ 559.92 mol
Next, we want to find the volume of the gas at STP. At STP, the temperature is 0°C (273.15 K) and the pressure is 1 atm.
Now we can use the ideal gas law again to find the volume:
V(STP) = (n * R * T(STP)) / P(STP)
Substituting the values:
V(STP) = (559.92 mol * 0.0821 L·atm/(mol·K) * 273.15 K) / (1 atm)
Simplifying the equation:
V(STP) ≈ 12445.942 L
To express the volume in liters to two significant digits, we round the result:
V(STP) ≈ 12000 L
Therefore, the volume of the gas at STP would be approximately 12000 liters.