Chlorine gas, Cl2, is widely used to purify municipal water supplies and treat swimming pool waters. If it occupies 23.5 cm^3 at STP, what volume will it occupy at -270. degrees C and a pressure of 1520 Torr?
Vf = 1.29 x 10^-4
I obtained 0.129 cc. If your answer is in L that is ok.
Yes, Liters. sorry
OK. You should always include units.
To find the volume of chlorine gas at -270 degrees C and a pressure of 1520 Torr, we can use the combined gas law equation, which relates the initial and final conditions of pressure and temperature to the initial and final volumes of a gas.
The combined gas law equation is: (P1 * V1) / (T1) = (P2 * V2) / (T2)
Here, P1, V1, and T1 represent the initial conditions and P2, V2, and T2 represent the final conditions.
Let's insert the values into the equation:
P1 = pressure at STP = 1 atmosphere = 760 Torr
V1 = volume at STP = 23.5 cm^3
T1 = temperature at STP = 0 degrees C = 273.15 K
P2 = 1520 Torr (given)
T2 = -270 degrees C + 273.15 = 3.15 K (converted to Kelvin)
Now we can solve for V2:
(760 Torr * 23.5 cm^3) / (273.15 K) = (1520 Torr * V2) / (3.15 K)
Simplifying the equation:
V2 = ((760 Torr * 23.5 cm^3 * 3.15 K) / (273.15 K * 1520 Torr))
V2 ≈ 1.29 x 10^-4 cm^3
So, the volume of chlorine gas at -270 degrees C and a pressure of 1520 Torr is approximately 1.29 x 10^-4 cm^3.