if 16.5 liters of gas X is at 398 torr and -90.5 degrees celsius, what would be the volume of this same gas if the pressure is changed to 3.1 atmospheres and the temperature is changed to 399K
To solve this problem, we can use the combined gas law equation, which relates the initial and final conditions of a gas when changes are made to pressure, volume, and temperature. The equation is as follows:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
Where:
P1 and P2 are the initial and final pressures, respectively;
V1 and V2 are the initial and final volumes, respectively;
T1 and T2 are the initial and final temperatures, respectively.
Let's plug in the given values into the equation:
P1 = 398 torr
V1 = 16.5 liters
T1 = -90.5 degrees Celsius
P2 = 3.1 atmospheres
T2 = 399K
Note: To convert Celsius to Kelvin, we add 273.15 to the Celsius temperature.
First, let's convert the initial temperature to Kelvin:
T1 = -90.5 + 273.15 = 182.65K
Now, we can plug in the values into the combined gas law equation:
(398 torr * 16.5 liters) / (182.65K) = (3.1 atmospheres * V2) / (399K)
Next, rearrange the equation to solve for V2:
V2 = ((398 torr * 16.5 liters) / (182.65K)) * (399K / (3.1 atmospheres))
Now, calculate the volume V2:
V2 = (6579 torr * liters) / (3.1 atmospheres)
Note: torr and atmospheres are equivalent units, so the units will cancel out.
V2 = 2122.9 liters
Therefore, the volume of the gas X would be approximately 2122.9 liters when the pressure is changed to 3.1 atmospheres and the temperature is changed to 399K.