Question

An ideal gas is provided in 35.0-L cylinders at a pressure of 16.0 atm. All the gas in three (3) cylinders at 20.5 °C is transferred to a rigid steel tank, then warmed to 31.5

°C. At 31.5 °C, the pressure in the tank is 1.25 atm. What is the volume of this steel tank?

Answer 1

The volume of the steel tank is calculated to be approximately 12,900 L after applying the combined gas law to the initial and final states of the ideal gas, taking into account the changes in temperature and pressure.

Step-by-step explanation:

To solve for the volume of the steel tank that the gas is being transferred into, we can use the combined gas law which relates pressure, volume, and temperature of a gas. The combined gas law is stated as (P1 * V1) / T1 = (P2 * V2) / T2, where P represents pressure, V represents volume, and T represents temperature in Kelvins. We need to make sure all units are consistent and temperatures are converted to Kelvins.

First, we'll convert the temperatures from Celsius to Kelvin.

• 
  
• Initial temperature (T1): 20.5°C + 273.15 = 293.65 K
• 
  
• Final temperature (T2): 31.5°C + 273.15 = 304.65 K

Since the same gas is being used, we can equate the product of pressure and volume for the initial and final states:

Initial combined pressure for three cylinders: 3 * 16.0 atm = 48.0 atm

Initial combined volume for three cylinders: 3 * 35.0 L = 105.0 L

Now we'll apply the combined gas law:
((48.0 atm * 105.0 L) / 293.65 K) = ((1.25 atm * V2) / 304.65 K)

By isolating V2 (the volume of the steel tank) and solving for it, we get:

V2 = (48.0 atm * 105.0 L * 304.65 K) / (1.25 atm * 293.65 K)

V2 = 12,968.4 L (This value needs to be rounded to the correct number of significant figures, based on given data, typically three significant figures)

V2 ≈ 12,900 L, which is the volume of the steel tank.