The average speed of a nitrogen molecule in air is proportional to the square root of the temperature in kelvins. If the average speed is 475 m/s on a warm summer day (temperature = 300.0 kelvins), what is the average speed on a cold winter day (250.0 kelvins)?
To find the average speed of a nitrogen molecule on a cold winter day, we can use the proportionality between the average speed and the square root of the temperature.
Let's denote V1 as the average speed on the warm summer day (300.0 kelvins) and T1 as the temperature on the warm summer day. Similarly, let's denote V2 as the unknown average speed on the cold winter day and T2 as the temperature on the cold winter day.
According to the given information, we have:
V1 = 475 m/s
T1 = 300.0 K
We can set up a proportion using the square roots of the temperatures and average speeds:
√T1 / V1 = √T2 / V2
Substituting the known values:
√300.0 / 475 = √T2 / V2
To solve for V2, we can rearrange the equation:
√T2 = (√300.0 / 475) * V2
Square both sides to isolate T2:
T2 = ((√300.0 / 475) * V2)^2
Now we can plug in the values and calculate T2:
T2 = ((√300.0 / 475) * V2)^2
T2 = ((√300.0 / 475) * V2) * ((√300.0 / 475) * V2)
T2 = ( √(300.0 * 300.0 ) / (475 * 475) ) * V2 * V2
T2 = ( 300.0 * 300.0 / (475 * 475) ) * V2 * V2
Calculating the numerical value:
T2 = (90000 / (225625)) * V2 * V2
T2 = 0.39978 * V2 * V2
Now we can substitute the value of T2 into the equation and solve for V2:
V2 = √T2
V2 = √(0.39978 * V2 * V2)
Simplifying:
V2 = √(0.39978) * √(V2 * V2)
V2 = 0.632139 * V2
To isolate V2, divide both sides of the equation by 0.632139:
V2 / 0.632139 = V2
Simplifying further:
1 / 0.632139 = 1
Therefore, the average speed of a nitrogen molecule on a cold winter day (250.0 kelvins) would be the same as the average speed on a warm summer day, which is 475 m/s.