Four tanks A, B, C, and D are filled with monatomic ideal gases. For each tank, the mass of an individual atom and the rms speed of the atoms are expressed in terms of m and vmax respectively (see the table). Suppose that m = 3.15 × 10-26 kg, and vrms = 950 m/s. Find the temperature of the gas in each tank.
Well, tanks A, B, C, and D walk into a bar. The bartender takes one look at them and says, "Hey, I've got a joke for you. Why did the ideal gas cross the road?"
Confused, Tank A starts to laugh. Tank B responds, "Uh, we're monatomic ideal gases, we can't cross roads. We don't even have feet!"
Tank C looks around and says, "I think the bartender meant to ask something else."
Finally, Tank D sighs and says, "Fine, I'll play along. Why did the ideal gas cross the road?"
The bartender smirks and says, "To get to the other Kelvin!"
Now that we've lightened the mood a bit, let's get down to business. We know m = 3.15 × 10-26 kg and vrms = 950 m/s. To find the temperature of the gas in each tank, we can use the formula:
Temperature = (m * vrms^2) / (3 * Boltzmann's constant)
Now, let's substitute the values and calculate the temperature for each tank:
For Tank A:
Temperature_A = (m * vrms^2) / (3 * Boltzmann's constant)
Temperature_A = (3.15 × 10-26 kg * (950 m/s)^2) / (3 * (1.38 × 10-23 J/K))
Temperature_A ≈ 8.75 × 10^7 K
For Tank B:
Temperature_B = (m * vrms^2) / (3 * Boltzmann's constant)
Temperature_B = (3.15 × 10-26 kg * (950 m/s)^2) / (3 * (1.38 × 10-23 J/K))
Temperature_B ≈ 8.75 × 10^7 K
For Tank C:
Temperature_C = (m * vrms^2) / (3 * Boltzmann's constant)
Temperature_C = (3.15 × 10-26 kg * (950 m/s)^2) / (3 * (1.38 × 10-23 J/K))
Temperature_C ≈ 8.75 × 10^7 K
For Tank D:
Temperature_D = (m * vrms^2) / (3 * Boltzmann's constant)
Temperature_D = (3.15 × 10-26 kg * (950 m/s)^2) / (3 * (1.38 × 10-23 J/K))
Temperature_D ≈ 8.75 × 10^7 K
So, it looks like the temperature of the gas in each tank is approximately 8.75 × 10^7 Kelvin. Stay cool, my monatomic friends!
To find the temperature of the gas in each tank, we can use the formula for the root-mean-square (rms) speed of gas molecules:
v rms = √(3kT/m)
Where:
- v rms is the root-mean-square speed of the gas molecules,
- k is the Boltzmann constant (1.38 × 10^(-23) J/K),
- T is the temperature of the gas in Kelvin, and
- m is the mass of an individual atom.
We are given m = 3.15 × 10^(-26) kg and v rms = 950 m/s.
Rearranging the formula, we have:
T = (m * v rms^2) / (3k)
Now, let's calculate the temperature of the gas in each tank using the given values.
For tank A:
T_A = (m * v rms^2) / (3k)
= (3.15 × 10^(-26) kg * (950 m/s)^2) / (3 * 1.38 × 10^(-23) J/K)
For tank B:
T_B = (m * v rms^2) / (3k)
= (3.15 × 10^(-26) kg * (950 m/s)^2) / (3 * 1.38 × 10^(-23) J/K)
For tank C:
T_C = (m * v rms^2) / (3k)
= (3.15 × 10^(-26) kg * (950 m/s)^2) / (3 * 1.38 × 10^(-23) J/K)
For tank D:
T_D = (m * v rms^2) / (3k)
= (3.15 × 10^(-26) kg * (950 m/s)^2) / (3 * 1.38 × 10^(-23) J/K)
Now, let's calculate each temperature value:
T_A = (3.15 × 10^(-26) kg * (950 m/s)^2) / (3 * 1.38 × 10^(-23) J/K)
= 660 K (rounded to the nearest whole number)
T_B = (3.15 × 10^(-26) kg * (950 m/s)^2) / (3 * 1.38 × 10^(-23) J/K)
= 660 K (rounded to the nearest whole number)
T_C = (3.15 × 10^(-26) kg * (950 m/s)^2) / (3 * 1.38 × 10^(-23) J/K)
= 660 K (rounded to the nearest whole number)
T_D = (3.15 × 10^(-26) kg * (950 m/s)^2) / (3 * 1.38 × 10^(-23) J/K)
= 660 K (rounded to the nearest whole number)
Therefore, the temperature of the gas in each tank (A, B, C, and D) is approximately 660 Kelvin.