Question

Standing waves of frequency 5.0 kHz are produced in a tube filled with oxygen at 300 K. The separation between the consecutive nodes is 3.3 cm. Calculate the specific heat capacities Cp and Cv of the gas.

Answer 1

Standing waves of frequency 5.0 kHz are produced in a tube filled with oxygen at 300 K. The separation between the consecutive nodes is 3.3 cm. The specific heat capacities Cp and Cv of the gas are: Cv = 20.7 J/mol K, Cp = 29 J/mol K.

• Given,
• Standing waves of frequency = f = 5 kHz = 500 Hz
• Temperature of oxygen = T = 300 K
• The separation between the consecutive nodes = l/2 = 3.3 cm
• ∴ l = 6.6 cm
• we have,
• v = fl = 500 * 6.6 = 330 m/s
• Formula
• $v = \sqrt{\dfrac{yRT}{M} } \\\\v^2= \dfrac{yRT}{M}\\\\330^2 = \dfrac{2490y}{32 * 10^{-3}}\\\\108900 = \dfrac{y * 8.3 * 300 }{32 * 10^{-3}}\\\\y = 1.3995$
• Now,
• The specific heat at constant volume,
• Cv = R / (y-1)
• $C_v=\dfrac{8.3}{1.3995}\\\\C_v=20.7J/mol K$
• The specific heat at constant pressure,
• Cp = Cv + R
• Cp=20.7 + 8.3
• Cp = 29 J/mol K

Answer 2

$C_{P}=29.0 \mathrm{J} / \mathrm{mol}-\mathrm{K}$, $C_{v}=$$20.7 \mathrm{J} \mathrm{mol}-\mathrm{K}$

Explanation:

Step 1:

Standing wave frequency , $f=5 \times 10^{3} \mathrm{Hz}$

Oxygen Temperature, $T=300 \mathrm{K}$

Separation of nodes in tunnel of the Kundt  

Step 2:

we know that

$\frac{l}{2}=\text { node separation }=3.3 \mathrm{cm}$

$\therefore l=6.6 \times 10^{-2} \mathrm{m}$

$\mathrm{v}=\mathrm{fl}=5 \times 10^{3} \times 6.6 \times 10^{-2}$

$=(66 \times 5)=330 \frac{\mathrm{m}}{\mathrm{s}}$

$\mathrm{v}=\sqrt{\frac{\gamma}{\mathrm{RTM}}}$

$\mathrm{v}^{2}=\frac{\gamma}{\mathrm{RTM}}$

$\gamma=\frac{32330 \times 330 \times 32}{8.3 \times 300 \times 100}$

$=1.3995$

Step 3:

Constant volume at specific heat , $C_{v}=\frac{R}{\gamma-1}=\frac{8.3}{0.3995}$

= $20.7 \mathrm{J} \mathrm{mol}-\mathrm{K}$

Constant pressure at specific heat ,  

$C_{P}=C_{v}+\mathrm{R}$

$C_{P}=20.7+8.3$

$C_{P}=29.0 \mathrm{J} / \mathrm{mol}-\mathrm{K}$