Final answer:
The new volume of a 20.0 L sample of gas at STP when the pressure is increased to 5.0 atm and the temperature to 100 degrees Celsius can be calculated using the Combined Gas Law and will be approximately 5.522 L.
Step-by-step explanation:
If a 20.0 L sample of gas originally at STP is increased to a pressure of 5.0 ATM and 100 degrees Celsius, we can find the new volume of the gas by using the Combined Gas Law, which is given by:
P1 * V1 / T1 = P2 * V2 / T2,
where P1 and V1 are the original pressure and volume, T1 is the original temperature, P2 and V2 are the final pressure and volume, and T2 is the final temperature.
Since the conditions are originally at STP (Standard Temperature and Pressure), we take P1 as 1 atm, V1 as 20.0 L, and T1 as 273 K (0 degrees Celsius). The final pressure P2 is given as 5.0 atm, and the final temperature T2 needs to be converted to Kelvin, which is 100 degrees Celsius + 273 = 373 K.
Now we can plug the values into the Combined Gas Law and solve for V2:
(1 atm * 20.0 L) / 273 K = (5.0 atm * V2) / 373 K
20.0 L * 373 K = 273 K * 5.0 atm * V2
V2 = (20.0 L * 373 K) / (273 K * 5.0 atm)
V2 = 5.522 L
So, the new volume of the gas at 5.0 atm and 100 degrees Celsius will be approximately 5.522 L.