How many times a diatomic gas should be expanded adiabatically so as to reduce the root mean square velocity to half A) 64 B) 32 C) 16 D) 8 Answer is B. Please explain.?

Question

Answer

Answer

Option B is a correct option
if we put the correct formula of mrms
we can find option no. B

Answer

To solve this problem, we need to understand the relationship between the root mean square (RMS) velocity of a diatomic gas and its expansion in an adiabatic process.

Adiabatic expansion is a thermodynamic process in which there is no exchange of heat between the system and its surroundings. In such a process, the total energy of the gas remains constant, but its volume and temperature change.

The RMS velocity of a gas is given by the equation:

v = √(3RT/M)

where v is the RMS velocity, R is the ideal gas constant, T is the temperature, and M is the molar mass of the gas.

Let's assume that the initial RMS velocity of the diatomic gas is v_0, and we want to reduce it to half, which is v_0/2.

Now, let's consider the adiabatic expansion process. During adiabatic expansion, the temperature and volume of the gas change, but the total energy remains constant.

In an adiabatic process, the relationship between temperature and volume can be described by the equation:

T1 * V1^(γ-1) = T2 * V2^(γ-1)

where T1 and V1 are the initial temperature and volume, respectively, T2 and V2 are the final temperature and volume, and γ is the heat capacity ratio for the gas.

For a diatomic gas, γ = 7/5.

Now, let's consider the first expansion. After the first expansion, the volume of the gas increases, and the temperature decreases. Let's denote the new temperature and volume as T1 and V1, respectively.

Since the total energy of the gas remains constant, we can write:

v_0 = √(3RT0/M)

v_0/2 = √(3RT1/M)

Dividing the second equation by the first equation, we get:

(√T1/√T0) = 1/√2

Squaring both sides of the equation, we get:

T1/T0 = 1/2

Using this relationship in the adiabatic expansion equation, we can write:

T0 * V0^(γ-1) = T1 * V1^(γ-1)

T0 * V0^(γ-1) = (1/2) * V1^(γ-1)

Since T1/T0 = 1/2, we can simplify the equation to:

V0^(γ-1) = V1^(γ-1)

Taking the γ-1 root of both sides of the equation, we get:

V0 = V1

This implies that the volume remains the same after the first expansion.

Since the volume remains the same, but the temperature decreases, the RMS velocity of the gas also decreases.

To reduce the RMS velocity to half, we need to repeat the adiabatic expansion process multiple times until the RMS velocity becomes v_0/2.

Since the volume remains the same after each expansion, the temperature will decrease by a factor of 2 each time.

Therefore, the number of expansions required to reduce the RMS velocity to half is equal to the logarithm of 2 to the base 2.

log2(2) = 1

log2(4) = 2

log2(8) = 3

...

log2(

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