What is the final volume of the gas in the following problem?
If I have 21 liters of gas held at a pressure of 78 atm and a temperature of 900 K, what will be the volume of the gas if I decrease the pressure to 45 atm and decrease the temperature to 750K?
This problem can be solved using the combined gas law:
P1V1/T1 = P2V2/T2
Initial conditions are:
P1 = 78 atm
V1 = 21 l
T1 = 900 K
P2 = 45 atm
V2 = ?
T2 = 750 K
Substituting your values, and solving for V2 you get:
V2 = 78 * 21 * 750/ (45 * 900) = 30.33 L
The problem states that there are initially 21 liters of a gas at a pressure of 78 atm. and the temperature of the gas is 900 K. The pressure is decreased to 45 atm. and the temperature is decreased to 750 K. The new volume as a result of the changes in the pressure and temperature has to be found.
This can be done using the ideal gas law. We have P*V = n*R*T, where P is the pressure, V is the volume, n is the amount of substance, R is a constant and T is the temperature. As the amount of the substance making up the gas n is a constant, the equation can be written as n*R = P*V/ T
Now the initial values and the final values that are known are substituted. Let the final volume be V.
21*78/900 = V*45/750
=> V = 21*78*750/900*45
=> V = 91/3 L
The required volume of the gas after the changes in pressure and temperature are made is 30.33 L.