A gaseous mixture enclosed in a vessel of volume V consists of one gram mole of gas A with gamma = C

Q: Why do we use the formula $\frac{n_1+n_2}{\gamma_{mix}-1} = \frac{n_1}{\gamma_1-1} + \frac{n_2}{\gamma_2-1}$?

This formula is derived from the conservation of internal energy. The total internal energy of the mixture is the sum of the internal energies of the individual gases: $U_{mix} = U_A + U_B$. Since $U = \frac{nRT}{\gamma-1}$, substituting this gives the relation for the mixture's gamma.

Q: Why do we need to calculate the molar mass of the mixture?

The speed of sound formula $v = \sqrt{\frac{\gamma RT}{M}}$ requires the molar mass of the specific gas. For a mixture, it behaves as a single gas with an effective molar mass, which is the total mass divided by the total number of moles.

Q: Can we just take the average of the speeds of sound of the two gases?

No! The speed of sound depends non-linearly on the mixture's effective gamma and effective molar mass. You must calculate these effective properties first and then plug them into the formula.

Q: Why did we use differentiation for part (c)?

When dealing with very small changes (like a 1 K increase in a 300 K room, which is ~0.33%), using calculus (differentials) provides a highly accurate and much faster way to find percentage changes compared to calculating the new speed and finding the difference.

Q: What exactly is adiabatic compressibility?

Compressibility measures how much a substance shrinks in volume when pressure is applied. It's the inverse of the Bulk Modulus. 'Adiabatic' just means this compression happens so fast (like in a sound wave) that no heat is exchanged with the surroundings.

Question

A gaseous mixture enclosed in a vessel of volume $V$ consists of one gram mole of gas $A$ with $\gamma = C_p/C_V = 5/3$ and another gas $B$ with $\gamma = 7/5$ at a certain temperature $T$. The gram molecular weights of the gases $A$ and $B$ are 4 and 32 respectively. The gases $A$ and $B$ do not react with each other and are assumed to be ideal. The gaseous mixture follows the equation $pV^{19/13} = 	ext{Constant}$, in adiabatic process. (1995)
(a) Find the number of gram moles of the gas $B$ in the gaseous mixture.
(b) Compute the speed of sound in the gaseous mixture at 300 K.
(c) If $T$ is raised by 1 K from 300 K, find the percentage change in the speed of sound in the gaseous mixture.
(d) The mixture is compressed adiabatically to 1/5 of its initial volume $V$. Find the change in its adiabatic compressibility in terms of the given quantities.

Sigma Percentile · Accepted Answer

This problem is a beautiful symphony of thermodynamics, blending the kinetic theory of gases, the physics of sound, error analysis, and the mechanics of adiabatic compression. Let's break down this multi-layered challenge step-by-step.

### The Mystery of the Missing Moles

We are given a vessel containing a mixture of two ideal gases, $A$ and $B$. We know everything about Gas $A$: it has $n_A = 1$ mole, an adiabatic exponent $\gamma_A = 5/3$, and a molar mass $M_A = 4$ g/mol. For Gas $B$, we know $\gamma_B = 7/5$ and $M_B = 32$ g/mol, but its number of moles, $n_B$, is unknown.

However, we are given a crucial clue: the mixture undergoes an adiabatic process following the equation $pV^{19/13} = 	ext{Constant}$. This immediately tells us that the effective adiabatic exponent of the mixture is $\gamma_{mix} = 19/13$.

To find the unknown moles of Gas $B$, we use the principle of conservation of internal energy, which gives us the formula for the equivalent $\gamma$ of a gaseous mixture:

$$ \frac{n_1 + n_2}{\gamma_{mix} - 1} = \frac{n_1}{\gamma_1 - 1} + \frac{n_2}{\gamma_2 - 1} $$

Substituting the known values:

$$ \frac{1 + n}{\frac{19}{13} - 1} = \frac{1}{\frac{5}{3} - 1} + \frac{n}{\frac{7}{5} - 1} $$

Simplifying the denominators:

$$ \frac{13(1 + n)}{6} = \frac{3}{2} + \frac{5n}{2} $$

Multiplying the entire equation by 6 to clear the fractions:

$$ 13 + 13n = 9 + 15n $$

$$ 2n = 4 \implies n = 2 	ext{ moles} $$

So, there are exactly 2 moles of Gas $B$ in the mixture.

### The Speed of Sound in a Crowd

Next, we need to compute the speed of sound in this mixture at $T = 300$ K. The speed of sound in an ideal gas is given by the Laplace-Newton formula:

$$ v = \sqrt{\frac{\gamma_{mix} R T}{M_{mix}}} $$

Before we can use this, we must find the effective molar mass of the mixture, $M_{mix}$. This is simply the total mass of the mixture divided by the total number of moles:

$$ M_{mix} = \frac{n_A M_A + n_B M_B}{n_A + n_B} $$

$$ M_{mix} = \frac{1(4) + 2(32)}{1 + 2} = \frac{68}{3} = 22.67 	ext{ g/mol} $$

**Crucial Step:** Always convert molar mass to standard SI units (kg/mol) before plugging it into physics formulas!

$$ M_{mix} = 22.67 	imes 10^{-3} 	ext{ kg/mol} $$

Now, substitute all values into the speed of sound formula:

$$ v = \sqrt{\frac{\left(\frac{19}{13}ight) (8.31) (300)}{22.67 	imes 10^{-3}}} \approx 401 	ext{ m/s} $$

### A Tiny Nudge in Temperature

If the temperature is raised by just 1 K from 300 K, what is the percentage change in the speed of sound? Because the change is very small ($\Delta T \ll T$), we can use the powerful tool of differentials.

We know that $v \propto \sqrt{T}$, which can be written as $v = K T^{1/2}$. Taking the natural logarithm on both sides:

$$ \ln v = \ln K + \frac{1}{2} \ln T $$

Differentiating this expression gives the fractional change:

$$ \frac{dv}{v} = \frac{1}{2} \frac{dT}{T} $$

To find the percentage change, multiply by 100:

$$ \% 	ext{ change} = \frac{dv}{v} 	imes 100 = \frac{1}{2} \left(\frac{dT}{T}ight) 	imes 100 $$

Substitute $dT = 1$ K and $T = 300$ K:

$$ \% 	ext{ change} = \frac{1}{2} \left(\frac{1}{300}ight) 	imes 100 = \frac{1}{6} \% \approx 0.167 \% $$

### The Squeeze: Adiabatic Compressibility

Finally, the mixture is compressed adiabatically to $1/5$ of its initial volume $V$. We need to find the change in its adiabatic compressibility.

Adiabatic compressibility, $\beta$, is defined as the reciprocal of the adiabatic bulk modulus ($B_{adi}$):

$$ \beta = \frac{1}{B_{adi}} = \frac{1}{\gamma p} $$

The change in compressibility is:

$$ \Delta \beta = \beta_f - \beta_i = \frac{1}{\gamma p_f} - \frac{1}{\gamma p_i} $$

To find the final pressure $p_f$, we use the adiabatic relation $p_i V_i^\gamma = p_f V_f^\gamma$:

$$ p_f = p_i \left(\frac{V_i}{V_f}ight)^\gamma $$

Since $V_f = V_i / 5$, the ratio $V_i / V_f = 5$. Thus, $p_f = p_i (5)^\gamma$.

Substitute this back into the $\Delta \beta$ equation:

$$ \Delta \beta = \frac{1}{\gamma p_i (5)^\gamma} - \frac{1}{\gamma p_i} = \frac{1}{\gamma p_i} \left( 5^{-\gamma} - 1 ight) $$

From the ideal gas law, the initial pressure is $p_i = \frac{n_{total} R T}{V} = \frac{3 R T}{V}$. Substituting this in:

$$ \Delta \beta = \frac{V}{3 \gamma R T} \left( 5^{-\gamma} - 1 ight) $$

Now, plug in the numerical values: $\gamma = 19/13$, $R = 8.31$, and $T = 300$:

$$ \Delta \beta = \frac{V}{3 \left(\frac{19}{13}ight) (8.31) (300)} \left( 5^{-19/13} - 1 ight) $$

$$ \Delta \beta = \frac{V}{10933.6} \left( 0.095 - 1 ight) $$

$$ \Delta \beta \approx -8.27 	imes 10^{-5} V $$

The negative sign indicates that as the gas is compressed, its pressure increases, making it harder to compress further (its compressibility decreases).

Visualized Solution

$n_{A} = 1, γ_{A} = 5/3, M_{A} = 4$

$\frac{n _{1} + n _{2}}{γ _{mi x} - 1} = \frac{n _{1}}{γ _{1} - 1} + \frac{n _{2}}{γ _{2} - 1}$

$\frac{1 + n}{\frac{19}{13} - 1} = \frac{1}{\frac{5}{3} - 1} + \frac{n}{\frac{7}{5} - 1}$

$v = \frac{γ _{mi x} R T}{M _{mi x}}$

$M_{mi x} = 22.67 g/mol$

$v \propto T$

$\frac{d v}{v} \times 100 = \frac{1}{2} (\frac{d T}{T}) \times 100$

$β = \frac{1}{B _{a d i}} = \frac{1}{γ p}$

$p_{i} V_{i}^{γ} = p_{f} V_{f}^{γ}$

$Δ β = \frac{1}{γ p _{i}} (5^{- γ} - 1)$

$Δ β \approx - 8.27 \times 1 0^{- 5} V$

The Sigma Insight: Thermodynamic Processes

The Mystery of the Missing Moles

The Speed of Sound in a Crowd

A Tiny Nudge in Temperature

The Squeeze

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } nA​=1,γA​=5/3,MA​=4

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } γmix​−1n1​+n2​​=γ1​−1n1​​+γ2​−1n2​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 1319​−11+n​=35​−11​+57​−1n​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } v=Mmix​γmix​RT​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Mmix​=22.67 g/mol

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } v∝T​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } vdv​×100=21​(TdT​)×100

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } β=Badi​1​=γp1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } pi​Viγ​=pf​Vfγ​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Δβ=γpi​1​(5−γ−1)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Δβ≈−8.27×10−5V