At atmospheric pressure, a sample of O2 gas has a volume of 55mL at 27 degrees celcius. What volume in mL will the O2 gas occupy if the temperature is decreased 0 degrees C and the pressure is held constant?
First, note the correct spelling of celsius.
I assume you have a typo "if the temperature is decreased TO 0.....".etc.
Use (P1V1)/T1 = (P2V2)/T2
Don't forget to change T to Kelvin.
To solve this problem, you can use the combined gas law equation that relates the initial and final volumes and temperatures of a gas sample:
(P1 × V1) / T1 = (P2 × V2) / T2
Where:
P1 = Initial pressure
V1 = Initial volume
T1 = Initial temperature
P2 = Final pressure (constant pressure)
V2 = Final volume (what we need to find)
T2 = Final temperature
In this case, we know:
P1 = P2 (constant pressure)
V1 = 55 mL
T1 = 27 °C = 27 + 273 = 300 K
T2 = 0 °C = 0 + 273 = 273 K
Plugging these values into the equation, we can solve for V2:
(P1 × V1) / T1 = (P2 × V2) / T2
(P2 × V2) = (P1 × V1 × T2) / T1
V2 = (P1 × V1 × T2) / (P2 × T1)
Since P1 = P2, the pressure cancels out, and the equation becomes:
V2 = (V1 × T2) / T1
Substituting the known values:
V2 = (55 mL × 273 K) / 300 K
V2 ≈ 50.55 mL
Therefore, the O2 gas will occupy approximately 50.55 mL if the temperature is decreased to 0 °C while holding the pressure constant.
To find the volume of the O2 gas at the new temperature (0 degrees Celsius), we need to apply Charles's Law.
Charles's Law states that the volume of a gas is directly proportional to its temperature, assuming the pressure remains constant. Mathematically, it can be expressed as V1/T1 = V2/T2, where V1 and T1 are the initial volume and temperature, and V2 and T2 are the final volume and temperature.
Given:
Initial volume, V1 = 55 mL
Initial temperature, T1 = 27 degrees Celsius
Final temperature, T2 = 0 degrees Celsius
Constant pressure throughout
First, we need to convert the temperatures to Kelvin since Charles's Law is based on the Kelvin scale.
Convert T1 to Kelvin: T1(K) = T1(°C) + 273.15
T1(K) = 27 + 273.15
T1(K) = 300.15 K
Convert T2 to Kelvin: T2(K) = T2(°C) + 273.15
T2(K) = 0 + 273.15
T2(K) = 273.15 K
Now we can use Charles's Law equation to find V2:
V1/T1 = V2/T2
Substituting the values:
55 mL / 300.15 K = V2 / 273.15 K
Now, cross-multiply and solve for V2:
55 mL × 273.15 K = V2 × 300.15 K
V2 = (55 mL × 273.15 K) / 300.15 K
Calculating the result:
V2 = 50 mL (rounded to the nearest whole number)
Therefore, if the temperature is decreased to 0 degrees Celsius while maintaining constant pressure, the O2 gas will occupy approximately 50 mL.