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`        Two ideal gases have the same value of Cp/Cv = γ. What will be the value of this ratio for a mixture of the two gases in the ratio 1 : 2?`
3 years ago

396 Points
```										Sol. Considering two gases, in Gas(1) we have,
γ, Cp1 (Sp. Heat at const. ‘P’), Cv base 1 (Sp. Heat at const. ‘V’), n base 1 (No. of moles)
Cp base 1/Cv base 1 = γ & Cp base 1 - Cv base 1 = R
⇒ γCv base 1 - Cv base 1 = R ⇒ Cv base 1 (γ - 1) = R
⇒ Cv base 1 = R/γ – 1 & Cp base 1 = γR/γ - 1
In Gas(2) we have, γ, Cp2 (Sp. Heat at const. ‘P’), Cv base 2 (Sp. Heat at const. ‘V’), n base 2 (No. of moles)
Cp base 2/Cv base 2 = γ & Cp base 2 - Cv base 2 = R ⇒ γCv base 2 - Cv base 2 = R ⇒ Cv base 2 (γ - 1) = R ⇒ Cv base 2 = R/γ – 1 & Cp base 2 = γR/γ – 1
Given n base 1 : n base 2 = 1 :2
dU base 1 = nCv base 1 dT & dU base 2 = 2nCv base 2 dT = 3nCvdT
⇒ Cv = Cv base 1 + 2Cv base 2/3 = R/γ – 1 + 2R/γ – 1/3 = 3R/3(γ – 1) = R/γ – 1  …(1)
&Cp = γCv = γr/γ – 1  …(2)
So, Cp/Cv = γ [from (1) & (2)]

```
3 years ago
Jitender Pal
365 Points
```										Sol. . Considering two gases, in Gas(1) we have,
γ, Cp1 (Sp. Heat at const. ‘P’), Cv base 1 (Sp. Heat at const. ‘V’), n base 1 (No. of moles)
Cp base 1/Cv base 1 = γ & Cp base 1 - Cv base 1 = R
⇒ γCv base 1 - Cv base 1 = R ⇒ Cv base 1 (γ - 1) = R
⇒ Cv base 1 = R/γ – 1 & Cp base 1 = γR/γ - 1
In Gas(2) we have, γ, Cp2 (Sp. Heat at const. ‘P’), Cv base 2 (Sp. Heat at const. ‘V’), n base 2 (No. of moles)
Cp base 2/Cv base 2 = γ & Cp base 2 - Cv base 2 = R ⇒ γCv base 2 - Cv base 2 = R ⇒ Cv base 2 (γ - 1) = R ⇒ Cv base 2 = R/γ – 1 & Cp base 2 = γR/γ – 1
Given n base 1 : n base 2 = 1 :2
dU base 1 = nCv base 1 dT & dU base 2 = 2nCv base 2 dT = 3nCvdT
⇒ Cv = Cv base 1 + 2Cv base 2/3 = R/γ – 1 + 2R/γ – 1/3 = 3R/3(γ – 1) = R/γ – 1  …(1)k
&Cp = γCv = γr/γ – 1  …(2)
So, Cp/Cv = γ [from (1) & (2)]

```
3 years ago
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