Question

An ideal gas of density 1.7 × 10−3 g cm−3 at a pressure of 1.5 × 105 Pa is filled in a Kundt's tube. When the gas is resonated at a frequency of 3.0 kHz, nodes are formed at a separation of 6.0 cm. Calculate the molar heat capacities Cp and Cv of the gas.

Answer 1

Molar heat capacity Cp = 26 J/mol - K and

Cv = 17.7 J/mol-K

Explanation:

Given data:

Density of gas "ρ" =1.7×10−3 g/cm^3  = 1.7 k/gm^3

Pressure "P" =1.5×10^5 Pa

R = 8.3 J/mol−K,

f =  3.0 KHz

Node separation = l/2 = 6 cm

l = 12 cm = 12×10^−2 m

Therefore

V = fl=3×10^3×12×10^−2

= 360 m/s

Speed of sound "V" = √γp/ρ

γ  = V^2ρ/p

γ  = 360 x 360 x 1.7/ 1.5 x 10^5

Cv  = R/ γ -1

Cv  = 8.3 / 1.468 -1  = 8.3 / 0.468

Cv = 17.7 J/mol-K

Cp = γCv = 1.468 x 17.7

     = 26 J/mol - K

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Answer 2

The molar heat capacities Cp and Cv of the gas is $26 \mathrm{J} / \mathrm{mol}-\mathrm{K}$, $17.7 \frac{J}{m o l-K}$.

Explanation:

Step 1:

Given data,  

Ideal gas Density, $\rho=1.7 \times 10^{-3} \mathrm{g} / \mathrm{cm}^{3}$

$=1.7 \mathrm{k} / \mathrm{gm}^{3}$

Gas Pressure, $P=1.5 \times 10^{5} \mathrm{Pa}$

$R=8.3 \mathrm{J} / \mathrm{mol}-\mathrm{K}$

Gas Resonance frequency = $3.0 \mathrm{kHz}$

Separation of nodes in tunnel of the Kundt

$=\frac{l}{2}=6 \mathrm{cm}$

Step 2:

$\text { So, } I=2 \times \times 6=12 \mathrm{cm}=12 \times 10^{-2} \mathrm{m}$

$\text { So, } V=f l=3 \times 10^{3} \times 12 \times 10^{-2}$

$=360 \mathrm{m} / \mathrm{s}$    

Sound of speed  $V=\frac{\gamma p}{\rho}$

$V^{2}=\frac{\gamma p}{\rho}$

$\gamma=\frac{v^{2} \rho}{P}=\frac{(360)^{2} \times 1.7 \times 10^{-3}}{1.5 \times 5}$

$=1.4688$

Using  $C_{p}-C_{v}=R$

$\frac{C_{p}}{C_{v}}=\gamma$

Step 3:

We know that,  

$C_{v}=\frac{R}{\gamma-1}=\frac{8.3}{0.4688}$

$=17.7 \frac{J}{m o l-K}$

$C_{p}=R+C_{v}=8.3+17.7=26 \mathrm{J} / \mathrm{mol}-\mathrm{K}$