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  • Two moles of an ideal gas with  are mixed 3 moles of another ideal gas with <img alt="\frac{C_p}{C_v}=\frac{4}{3}" src="http:/

Two moles of an ideal gas with \frac{C_p}{C_v}=\frac{5}{3} are mixed 3 moles of another ideal gas with \frac{C_p}{C_v}=\frac{4}{3}. The value of \frac{C_p}{C_v} for the mixture is:-  
Option: 1 1.50
Option: 21.45
Option: 3 1.47    
Option: 4 1.42
 

Answers (1)

best_answer

For ideal gas:- C_p-C_v=R

For first case:-

\frac{C_{p1}}{C_{v1}}=\frac{5}{3} \ and \ C_{p1}-C_{v1}=R

C_{p1}=\frac{5}{3}{C_{v1}} \ and \ \frac{5}{3}{C_{v1}} -C_{v1}=R\Rightarrow \frac{2}{3}C_{v1}=R\Rightarrow C_{v1}=\frac{3}{2}R

So, C_{p1}=\frac{5}{2}R

For second case:-

\frac{C_{p2}}{C_{v2}}=\frac{4}{3} \ and \ C_{p2}-C_{v2}=R

C_{p2}=\frac{4}{3}{C_{v2}} \ and \ \frac{4}{3}{C_{v2}}-C_{v2}=R\Rightarrow C_{v2}=3R and C_{p2}=4R

Now, Y_{mix}=\frac{n_1C_{p1}+n_2C_{p2}}{n_1C_{v1}+n_2C_{v2}}=\frac{2\times\frac{5}{2}R+3\times4R}{2\times\frac{3}{2}R+3\times3R}= 1.417 = 1.42

So option (4) is correct. 

Posted by

Ritika Jonwal

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