Fifty liters of gas is kept at a temperature of 200 K and under pressure of 15 atm. The temperature of the gas is increased to 400 K. The pressure is decreased to 7.5 atm. What is the resulting volume of the gas?
P1V1/T1 = P2V2/T2
solve for V2
To solve this problem, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature
We are given the initial conditions:
Initial volume (V1) = 50 liters
Initial temperature (T1) = 200 K
Initial pressure (P1) = 15 atm
We are also given the final conditions:
Final temperature (T2) = 400 K
Final pressure (P2) = 7.5 atm
Since the number of moles (n) remains constant, we can rewrite the equation as:
P1V1/T1 = P2V2/T2
Now let's plug in the values:
15 atm * 50 L / 200 K = 7.5 atm * V2 / 400 K
Simplifying the equation:
750 atm·L / 200 K = 7.5 atm·V2 / 400 K
To solve for V2, we can cross multiply:
750 atm·L * 400 K = 7.5 atm * V2 * 200 K
Divide both sides of the equation by (7.5 atm * 200 K):
V2 = (750 atm·L * 400 K) / (7.5 atm * 200 K)
Cancel out the units:
V2 = 400 L
Therefore, the resulting volume of the gas is 400 liters.