Question 181: Thermodynamics & KTG

Question

Match the List I with List II

List - I

List - II

(A)
Triatomic rigid gas
(I)
$\frac{C_p}{C_v}=\frac{5}{3}$

(B)
Diatomic non-rigid gas
(II)
$\frac{C_p}{C_v}=\frac{7}{5}$

(C)
Monoatomic gas
(III)
$\frac{C_p}{C_v}=\frac{4}{3}$

(D)
Diatomic rigid gas
(IV)
$\frac{C_p}{C_v}=\frac{9}{7}$

Choose the correct answer from the options given below:

Accepted Answer

The explanation uses the equation for the heat capacity ratio, $ \gamma = 1 + \frac{2}{	ext{f}} $, where $ 	ext{f} $ is the degrees of freedom of the gas. Here's how it applies to different types of gases:
Triatomic Rigid Gas
:
Degrees of freedom ($ 	ext{f} $) = 6
$ \gamma = 1 + \frac{2}{6} = \frac{4}{3} $
Diatomic Non-Rigid Gas
:
Degrees of freedom ($ 	ext{f} $) = 7
$ \gamma = 1 + \frac{2}{7} = \frac{9}{7} $
Diatomic Rigid Gas
:
Degrees of freedom ($ 	ext{f} $) = 5
$ \gamma = 1 + \frac{2}{5} = \frac{7}{5} $
Monoatomic Gas
:
Degrees of freedom ($ 	ext{f} $) = 3
$ \gamma = 1 + \frac{2}{3} = \frac{5}{3} $
Therefore, when matching List I with List II, the correct pairing is:
A
(Triatomic rigid gas) -
III
$(\frac{4}{3})$
B
(Diatomic non-rigid gas) -
IV
$(\frac{9}{7})$
C
(Monoatomic gas) -
I
$(\frac{5}{3})$
D
(Diatomic rigid gas) -
II
$(\frac{7}{5})$
Thus, the correct answer from the options given is:
A-III, B-IV, C-I, D-II

Answer

A-III, B-IV, C-I, D-II

Answer

A-II, B-IV, C-I, D-III

Answer

A-IV, B-II, C-III, D-I

Answer

A-III, B-II, C-IV, D-I

Thermodynamics & KTG

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