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Karnataka Board PUCPUC Science Class 11

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.

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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 in Brief
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Solution

Frequency of standing waves, f = 5 × 103 Hz
Temperature of oxygen, T = 300 K
From Kundt's tube theory, we know that`"l" / 2` = node separation = 3.3 cm
∴ l = 6.6 × 10−2 m
Also,
v = fl = 5 × 103 × 6.6 × 10−2
= (66 × 5) = 330 m/s

`"v" = sqrt((gamma "R""T")/"M")`

`"v"^2=(gamma "R""T")/"M"`

`therefore  gamma = (330 xx 330 xx 32)/(8.3 xx 300 xx 100 ) = 1.3995`

Specific heat at constant volume , `"C"_"v" = "R"/(gamma -1) = 8.3/0.3995`

= 20.7 J mol -K

Specific heat at constant pressure, Cp = Cv + R
Cp = 20.7 + 8.3
Cp = 29.0 J/mol-K

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Chapter 27: Specific Heat Capacities of Gases - Exercises [Page 80]

APPEARS IN

HC Verma Concepts of Physics Volume 1 and 2 [English]
Chapter 27 Specific Heat Capacities of Gases
Exercises | Q 35 | Page 80

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