A gas has a volume of 1,400 milliliters at a temperature of 20.0 K and a pressure of 101.3 kPa. What will be the volume when the temperature is changed to 40.0 K and the pressure is changed to 50.65 kPa?
To solve this problem, we can use the ideal gas law equation:
PV = nRT
Where:
P = Pressure
V = Volume
n = Number of moles of gas
R = Ideal gas constant
T = Temperature
To solve for the final volume (V2), we can rearrange the equation as follows:
V1 / T1 = V2 / T2
Where:
V1 = Initial volume
T1 = Initial temperature
V2 = Final volume (what we need to find)
T2 = Final temperature
Now, let's plug in the given values:
V1 = 1,400 mL = 1.4 L (since 1 L = 1,000 mL)
T1 = 20.0 K
P1 = 101.3 kPa
T2 = 40.0 K
P2 = 50.65 kPa
Using the equation, we have:
V1 / T1 = V2 / T2
Plugging in the values:
1.4 / 20.0 = V2 / 40.0
Now, cross-multiply:
1.4 × 40.0 = V2 × 20.0
Simplifying:
56 = 20V2
Divide both sides by 20:
56 / 20 = V2
V2 = 2.8 L
Therefore, the volume will be 2.8 liters when the temperature is changed to 40.0 K and the pressure is changed to 50.65 kPa.