Question

Q17) One mole of mono atomic gas (g = 5/3) is mixed with one mole of diatomic gas (g = 7/5) what will be the value of g for the mixture? (a) 1.5 (b) 2.5​

Answer 1

Answer:

Correct option is a)

Explanation:

As we know,

C

v

=

2

3

RT;C

p

=

2

5

RT for monoatomic gas;

C

v

=

2

5

RT;C

p

=

2

7

RT for diatomic gas

Thus, for mixture of 1 mole each,

C

v

=

2

2

3

RT+

2

5

RT

=2RT and C

p

=

2

2

5

RT+

2

7

RT

=3RT

Therefore, C

p

/C

v

=

2RT

3RT

=1.5

Answer 2

$\huge \mathcal \pink{ \to \: answer}$

(a) 1.5

$\large \mathbb{ \: Question\: \rightarrow}$

One mole of mono atomic gas (g = 5/3) is mixed with one mole of diatomic gas (g = 7/5) what will be the value of g for the mixture?

$\large \mathbb{To \: \: \: \: Find \to}$

what will be the value of g for the mixture?

$\large \mathbb {Solution \to}$

Gas is monoatomic :

$\big \gamma 1= \frac{5}{3} \\ \\ Cv1 \: = \frac{3}{2} R$

Gas is diatomic :

$\gamma 2 = \frac{7}{5} \\ \\ Cv2= \frac{5}{2} R$

Cv (of the mixture)

$\large = \frac{n_1 Cv_{1}+ \: n_2 Cv_{2} \: }{ \: \: n_1 + n_2\: } \\ \\ \large = \frac{(1)( \frac{3}{2} R ) + (1)( \frac{5}{2} R)}{1 + 1} \\ \\ \large C _{p} \: (of \: the \: mixture) \\ \large = C _{v} \: + \: R \: = 31 \\ \\ \large \therefore \gamma \: mixture \: = \frac{C _{p}}{C _{v}} = \\ \large = \: \frac{3R}{2R} = 1.5$

therefore option (A) is right option

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