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A container is divided into two compartm...

A container is divided into two compartments by a removable partition as shown below :

In the first compartment `n_1` moles of ideal gas He is present in a volume `V_1` . In the second compartment, `n_2` moles of ideal gas Ne is present in a volume `V_2`. The temperature and pressure in both the compartments are T and P repectively. Assuming R is the gas constant. the total change is entropy upon removing the partition when the gases mix irreversibly is :

A

`n_1 Rl n (V_1)/(V_1 +V_2) + n_2 Rl n (V_2)/(v_1 +V_2)`

B

`n_1 Rl n (V_1 +V_2)/(V_1) +n_2 Rln (V_1 +V_2)/(V_2) `

C

`(n_1 +n_2) Rl n (n_1 V_1)/(n_2 V_2)`

D

`(n_1 +n_2 ) Rln (n_2 V_2)/(n_1 V_1)`

Text Solution

Verified by Experts

The correct Answer is:
B
Entropy change ` Delta S = n C_v l n`
`((T_2)/(T_1))+nRTlN ((V_2)/(V_1))`
since temperature is constant throughout process
He ` :Delta S = n_1 Rln ((V_1 + V_2)/(v_1))`
Ne :` Delta S = n_2 Rl n (V_1+V_2)/(V_2)`
total change in `(Delta S ) = n_1 R l n (V_1+V_2)/(V_1)) + n_2Rl n ((V_1 +V_2)/( V_2))`
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