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Two moles of helium are mixed with n moles of hydrogen.

If \frac{C_{p}}{C_{v}}= 3/2

for the mixture, then the value of n is :

  • Option 1)

    1

  • Option 2)

    3

  • Option 3)

    2

  • Option 4)

    3/2

 

Answers (1)

best_answer

As we have learned

Atomicity or adiabatic coefficient (gamma) -

\gamma =\frac{ C_{p}}{ C_{v}}

=1+\frac{2}{f}

 

- wherein

for Monoatomic gas     \gamma= \frac{5}{3}

for Diatomic gas         \gamma= \frac{7}{5}

for Triatomic gas        \gamma= \frac{4}{3}

 

c_{p}^{mix}= \frac{n_{1}cp_{1}+n_{2}cp_{2}}{n_{1}+n_{2}}=\frac{{2*5R/2}+n(7R/2)}{2+n}= \frac{10+7n}{2+ n}R

c_{v}^{mix}= \frac{n_{1}cp_{1}+n_{2}cp_{2}}{n_{1}+n_{2}}=\frac{{2*3R/2}+n(5R/2)}{2+n}= \frac{6+5n}{2+ n}R 

r^{mix}= \frac{3}{2}= \frac{c_{p}^{mix}}{c_{v}^{mix}}=\frac{10+7n}{6+5n}\Rightarrow 18+15n=20+14n

or n=2

 

 

 

 

 

 


Option 1)

1

This is incorrect

Option 2)

3

This is incorrect

Option 3)

2

This is correct

Option 4)

3/2

This is incorrect

Posted by

Avinash

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