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If "7 gm N"(2) is mixed with 20 gm Ar, t...

If `"7 gm N"_(2)` is mixed with 20 gm Ar, there `C_(p)//C_(v)` of mixture will be :

A

`(17)/(6)`

B

`(11)/(7)`

C

`(17)/(11)`

D

`(17)/(13)`

Text Solution

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The correct Answer is:
To solve the problem of finding the ratio \( \frac{C_p}{C_v} \) for a mixture of 7 gm of nitrogen (\( N_2 \)) and 20 gm of argon (\( Ar \)), we will follow these steps: ### Step 1: Determine the number of moles of each gas. 1. For nitrogen (\( N_2 \)): - Molar mass of \( N_2 = 28 \, \text{g/mol} \) - Number of moles of \( N_2 = \frac{7 \, \text{g}}{28 \, \text{g/mol}} = \frac{1}{4} \, \text{mol} \) 2. For argon (\( Ar \)): - Molar mass of \( Ar = 40 \, \text{g/mol} \) - Number of moles of \( Ar = \frac{20 \, \text{g}}{40 \, \text{g/mol}} = \frac{1}{2} \, \text{mol} \) ### Step 2: Calculate \( C_p \) and \( C_v \) for each gas. 1. For nitrogen (\( N_2 \)): - Degrees of freedom \( F = 5 \) (as it is diatomic) - \( C_p = \left( F + 1 \right) \frac{R}{2} = \left( 5 + 1 \right) \frac{R}{2} = 7 \frac{R}{2} \) - \( C_v = F \frac{R}{2} = 5 \frac{R}{2} \) 2. For argon (\( Ar \)): - Degrees of freedom \( F = 3 \) (as it is monoatomic) - \( C_p = \left( F + 1 \right) \frac{R}{2} = \left( 3 + 1 \right) \frac{R}{2} = 5 \frac{R}{2} \) - \( C_v = F \frac{R}{2} = 3 \frac{R}{2} \) ### Step 3: Calculate the average \( C_p \) and \( C_v \) for the mixture. 1. For the mixture: - \( C_{p, mix} = \frac{n_{N_2} C_{p, N_2} + n_{Ar} C_{p, Ar}}{n_{N_2} + n_{Ar}} \) - \( C_{v, mix} = \frac{n_{N_2} C_{v, N_2} + n_{Ar} C_{v, Ar}}{n_{N_2} + n_{Ar}} \) 2. Substituting the values: - \( C_{p, mix} = \frac{\left(\frac{1}{4}\right) \left(7 \frac{R}{2}\right) + \left(\frac{1}{2}\right) \left(5 \frac{R}{2}\right)}{\frac{1}{4} + \frac{1}{2}} \) - \( C_{p, mix} = \frac{\frac{7R}{8} + \frac{5R}{4}}{\frac{3}{4}} = \frac{\frac{7R}{8} + \frac{10R}{8}}{\frac{3}{4}} = \frac{\frac{17R}{8}}{\frac{3}{4}} = \frac{17R}{6} \) 3. For \( C_{v, mix} \): - \( C_{v, mix} = \frac{\left(\frac{1}{4}\right) \left(5 \frac{R}{2}\right) + \left(\frac{1}{2}\right) \left(3 \frac{R}{2}\right)}{\frac{1}{4} + \frac{1}{2}} \) - \( C_{v, mix} = \frac{\frac{5R}{8} + \frac{3R}{4}}{\frac{3}{4}} = \frac{\frac{5R}{8} + \frac{6R}{8}}{\frac{3}{4}} = \frac{\frac{11R}{8}}{\frac{3}{4}} = \frac{11R}{6} \) ### Step 4: Calculate the ratio \( \frac{C_p}{C_v} \). - \( \frac{C_p}{C_v} = \frac{\frac{17R}{6}}{\frac{11R}{6}} = \frac{17}{11} \) ### Final Answer: The ratio \( \frac{C_p}{C_v} \) of the mixture is \( \frac{17}{11} \). ---
To solve the problem of finding the ratio \( \frac{C_p}{C_v} \) for a mixture of 7 gm of nitrogen (\( N_2 \)) and 20 gm of argon (\( Ar \)), we will follow these steps: ### Step 1: Determine the number of moles of each gas. 1. For nitrogen (\( N_2 \)): - Molar mass of \( N_2 = 28 \, \text{g/mol} \) - Number of moles of \( N_2 = \frac{7 \, \text{g}}{28 \, \text{g/mol}} = \frac{1}{4} \, \text{mol} \) ...
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