A sealed cylinder contains a sample of ideal gas at a pressure of 2.0 atm. The rms
speed of the molecules is vO.
(a) If the rms speed is then reduced to 0.90 vO, what is the pressure of the gas?
(b) By what percentage does the speed of sound in the gas change?
To solve this problem, we can use the relationship between pressure and the rms speed of gas molecules. The rms speed of gas molecules is related to the temperature of the gas.
(a) Let's start by assuming that the temperature of the gas remains constant even after the rms speed is reduced to 0.90 vO. According to the ideal gas law, PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant, and T is the temperature.
Since the temperature remains constant, we can rewrite the ideal gas law as P1V = P2V, where P1 is the initial pressure, V is the initial volume, and P2 is the final pressure after the rms speed reduction.
From the given information, P1 = 2.0 atm and V remains constant. Let's label the final pressure as P2.
So, we have: 2.0 atm * V = P2 * V.
V cancels out from both sides of the equation, leaving us with P2 = 2.0 atm * 0.90.
Calculating this, we find that P2 = 1.8 atm.
Therefore, the pressure of the gas after the rms speed is reduced to 0.90 vO is 1.8 atm.
(b) To calculate the percentage change in the speed of sound in the gas, we need to find the initial and final speeds of sound.
The speed of sound in a gas is given by the formula: v = √(γRT/M), where v is the speed of sound, γ is the adiabatic constant (specific heat capacity ratio), R is the gas constant, T is the temperature, and M is the molar mass of the gas.
Since the temperature remains constant in this case, the speed of sound is directly proportional to the rms speed of the gas molecules.
Let's label the initial speed of sound as v1 and the final speed of sound as v2.
Given that the initial rms speed of the gas molecules is vO, we can equate it to v1. And we are asked to find the percentage change, so the formula is given by:
Percentage change = ((v2 - v1) / v1) * 100
Since the initial and final rms speeds are related to the speeds of sound, we can write:
v1 = √(γRT/M) and v2 = √(γRT/M) * 0.9.
Now, let's substitute these values into the percentage change formula:
Percentage change = ((√(γRT/M) * 0.9 - √(γRT/M)) / √(γRT/M)) * 100.
Since the γ, R, and M are constants and cancel out from both sides of the equation, we can simplify this to:
Percentage change = (0.9 - 1) * 100.
Calculating this, we find that the percentage change in the speed of sound in the gas is -10%.
Therefore, the speed of sound in the gas decreases by 10% when the rms speed of the molecules is reduced to 0.90 vO.
To solve this problem, we can use the relationship between the rms speed of gas molecules and pressure.
(a) To find the pressure of the gas when the rms speed is reduced to 0.90 vO, we can use the following formula:
P1 = P0 * (V1^2 / V0^2)
where P1 is the new pressure, P0 is the initial pressure, V1 is the new rms speed, and V0 is the initial rms speed.
Plugging in the values, we have:
P1 = 2.0 atm * (0.90 vO / vO)^2
P1 = 2.0 atm * 0.90^2
Calculating the expression, we find:
P1 = 2.0 atm * 0.81
P1 = 1.62 atm
So, the pressure of the gas when the rms speed is reduced to 0.90 vO is 1.62 atm.
(b) Now, let's calculate the percentage change in the speed of sound.
The speed of sound in a gas is given by the formula:
v_sound = sqrt(gamma * R * T)
where gamma is the adiabatic index (equal to 7/5 for a diatomic gas), R is the gas constant, and T is the temperature.
Since the temperature is constant, the change in the speed of sound will be proportional to the square root of the pressure change:
(delta v_sound / v_sound) * 100 = (delta P / P) * 100
To find the percentage change in the speed of sound, we can use the following formula:
(delta v_sound / v_sound) * 100 = ((P1 - P0) / P0) * 100
Plugging in the values, we have:
(delta v_sound / v_sound) * 100 = ((1.62 atm - 2.0 atm) / 2.0 atm) * 100
(delta v_sound / v_sound) * 100 = (-0.38 atm / 2.0 atm) * 100
(delta v_sound / v_sound) * 100 = -19%
Therefore, the speed of sound in the gas decreases by 19%.