If 21.0 L of gas is held at a pressure of 7.80 atm and a temperature of 900.0 K, what will be the volume of the gas if the pressure is decreased to 4.50 atm and temperature decreased to 750.0 K?
Shehs
To solve this problem, we can use the combined gas law, which relates the pressure, volume, and temperature of a gas. The combined gas law formula is:
(P1 × V1) / (T1) = (P2 × V2) / (T2)
Where:
P1 = initial pressure
V1 = initial volume
T1 = initial temperature
P2 = final pressure
V2 = final volume (what we need to find)
T2 = final temperature
Given:
P1 = 7.80 atm
V1 = 21.0 L
T1 = 900.0 K
P2 = 4.50 atm
T2 = 750.0 K
Now we can substitute the values into the formula and solve for V2:
(7.80 atm × 21.0 L) / (900.0 K) = (4.50 atm × V2) / (750.0 K)
To isolate V2, we can cross multiply and then divide:
(7.80 atm × 21.0 L × 750.0 K) = (4.50 atm × V2 × 900.0 K)
Multiplying across:
11,715 atm*L*K = 4,050 atm*V2*K
Dividing both sides by 4,050 atm*K:
11,715 atm*L*K / 4,050 atm*K = V2
Simplifying:
2.89 L = V2
Therefore, the volume of the gas will be approximately 2.89 L when the pressure is decreased to 4.50 atm and the temperature is decreased to 750.0 K.