Heat and Thermodynamics 6 Question 32
37. 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 $p V^{19 / 13}=c$ onstant, in adiabatic process.
$(1995,10 M)$
(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.
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Answer:
Correct Answer: 37. $800 K, 720 J$
Solution:
- (a) Number of moles of gas $A$ are $n _A=1$ (given) Let the number of moles of gas $B$ be $n _B=n$
For a mixture
$$ \frac{n _1+n _2}{\gamma-1}=\frac{n _1}{\gamma _1-1}+\frac{n _2}{\gamma _2-1} $$
Substituting the values in Eq. (i), we have
$$ \frac{1+n}{(19 / 13)-1}=\frac{1}{(5 / 3)-1}+\frac{n}{(7 / 5)-1} $$
Solving this, we get $n=2$.
(b) Molecular weight of the mixture will be given by,
$$ \begin{aligned} M & =\frac{n _A M _A+n _B M _B}{n _A+n _B}=\frac{(1)(4)+2(32)}{1+2} \\ M & =22.67 \end{aligned} $$
Speed of sound in a gas is given by
$$ v=\sqrt{\frac{\gamma R T}{M}} $$
Therefore, in the mixture of the gas
(c)
$$ \begin{aligned} & v=\sqrt{\frac{(19 / 13)(8.31)(300)}{22.67 \times 10^{-3}}} m / s \\ & v \approx 401 m / s \end{aligned} $$
$$ v \propto \sqrt{T} $$
or $\quad v=K T^{1 / 2}$
$\Rightarrow \frac{d v}{d T}=\frac{1}{2} K T^{-1 / 2} \Rightarrow d v=K\left(\frac{d T}{2 \sqrt{T}}\right)$
$\Rightarrow \frac{d v}{v}=\frac{K}{v} \cdot\left(\frac{d T}{2 \sqrt{T}}\right) \Rightarrow \frac{d v}{v}=\frac{1}{\sqrt{T}}\left(\frac{d T}{2 \sqrt{T}}\right)=\frac{1}{2}\left(\frac{d T}{T}\right)$
$\Rightarrow \frac{d v}{v} \times 100=\frac{1}{2}\left(\frac{d T}{T}\right) \times 100=\frac{1}{2}\left(\frac{1}{300}\right) \times 100$
$$ =0.167 $$
Therefore, percentage change in speed is $0.167 %$.
(d) Compressibility $=\frac{1}{\text { Bulk modulus }}=\beta$ (say)
Adiabatic bulk modulus is given by
$$ B _{\text {adi }}=\gamma p \quad\left(B=-\frac{d p}{d V / V}\right) $$
$\therefore$ Adiabatic compressibility will be given by
$$ \begin{aligned} & \beta _{\text {adi }}=\frac{1}{\gamma p} \text { and } \beta _{\text {adi }}^{\prime}=\frac{1}{\gamma p^{\prime}}=\frac{1}{\gamma p(5)^{\gamma}}\left(p V^{\gamma}=\text { constant }\right) \\ & {\left[p V^{\gamma}=p^{\prime}(V / 5)^{\gamma} \Rightarrow p^{\prime}=p(5)^{\gamma}\right]} \\ & \therefore \Delta \beta=\beta _{\text {adi }}^{\prime}-\beta _{\text {adi }}=\frac{-1}{\gamma p}\left[1-\left(\frac{1}{5}\right)^{\gamma}\right] \\ & =\frac{-V}{\gamma\left(n _A+n _B\right) R T}\left[1-\left(\frac{1}{5}\right)^{\gamma}\right] \quad\left(\because p=\frac{n R T}{V}\right) \\ & =\frac{-V}{\left(\frac{19}{13}\right)(1+2)(8.31)(300)}\left[1-\left(\frac{1}{5}\right)^{\frac{19}{13}}\right] \\ & \left(\gamma=\gamma _{\text {mixture }}=\frac{19}{13}\right) \\ & \Delta \beta=-8.27 \times 10^{-5} V \end{aligned} $$
