A gas is at 20°C. To what temperature must it be raised to triple the rms speed of its molecules?
Ans: 2364oC
I don't get how they got the answer. Please help!
You wanker. At least explain it for others who find this in the future
To determine the temperature at which the gas molecules' root-mean-square (rms) speed is tripled, we can use the following formula relating the rms speed (vrms) to temperature (T):
vrms = √(3kT/m)
where k is the Boltzmann constant (1.38 x 10^-23 J/K) and m is the mass of a single gas molecule.
To triple the rms speed, we need to increase it by a factor of 3. Let's call the initial temperature T1 and the final temperature T2.
We can write the ratio of the two rms speeds as:
vrms(T2) / vrms(T1) = 3
Substituting the formulas for vrms(T2) and vrms(T1), we get:
√(3kT2/m) / √(3kT1/m) = 3
Simplifying the equation by canceling out the common factors:
√(T2/T1) = 3
To isolate the temperature variable, we square both sides of the equation:
T2/T1 = 3^2
T2/T1 = 9
Now, rearrange the equation to solve for T2:
T2 = 9T1
Given that the gas is initially at 20°C (or 293 K), we can substitute T1 into the equation:
T2 = 9 * 293
T2 ≈ 2637 K
Converting the final temperature from Kelvin to Celsius:
T2 ≈ 2637 - 273
T2 ≈ 2364°C
Therefore, the gas must be raised to approximately 2364°C to triple the rms speed of its molecules.
To find the temperature at which the root mean square (rms) speed of the gas molecules triples, we can use the formula for the rms speed of a gas molecule given by the equation:
v(rms) = √(3kT/m)
where v(rms) is the rms speed, k is the Boltzmann constant (1.38 x 10^-23 J/K), T is the temperature (in Kelvin), and m is the mass of a gas molecule.
In order to triple the rms speed, we need to find the temperature at which v(rms) becomes three times its initial value. Let's denote the initial temperature as T1 and the final temperature as T2.
We can set up the equation as follows:
√(3kT2/m) = 3√(3kT1/m)
To find T2, we need to solve this equation for T2:
√(T2) = 3√(T1)
Squaring both sides of the equation, we get:
T2 = 9T1
Now, let's plug in the given initial temperature T1 = 20°C = 293K.
T2 = 9 * 293K
T2 = 2637K
Finally, we need to convert the final temperature from Kelvin to Celsius:
Tfinal = T2 - 273.15
Tfinal = 2637K - 273.15
Tfinal ≈ 2364°C
Therefore, the gas needs to be raised to approximately 2364°C to triple the rms speed of its molecules.