A thermally insulated vessel is divided into two compartments A and B by a partition of insulating material- Compartment A has 0-1 mole of Helium at 427- C and 1 atm and compartment B has x mol of Neon at 127- C and 1 atm- If partition is removed and final temperature of gas is 227- C- then the value of 10x is - -Assume Cv-3R2 for both gases -

Heat gain by Neon = Heat loss by Helium ⇒ n_1C_v(500 400)=0.1× C_v(700 500) ⇒ n_1× 100=0.1× 200 nbsp; nbsp; nbsp; nbsp; n_1=0.2 ...

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Byju's Answer

Standard XII

Chemistry

Heat Capacity at Constant Volume

A thermally i...

Question

A thermally insulated vessel is divided into two compartments A and B by a partition of insulating material. Compartment A has 0.1 mole of Helium at $427^{\circ} C$ and 1 atm and compartment B has $x$ mol of Neon at $127^{\circ} C$ and 1 atm. If partition is removed and final temperature of gas is $227^{\circ} C$ , then the value of $10 x$ is . [Assume $C_{v} = \frac{3 R}{2}$ for both gases ]

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Solution

Heat gain by Neon = Heat loss by Helium
$\Rightarrow n_{1} C_{v} (500 - 400) = 0.1 \times C_{v} (700 - 500)$
$\Rightarrow n_{1} \times 100 = 0.1 \times 200$
$n_{1} = 0.2$

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Heat gain by Neon = Heat loss by Helium ⇒ n1Cv(500−400)=0.1×Cv(700−500) ⇒ n1×100=0.1×200 n1=0.2

Heat gain by Neon = Heat loss by Helium
$\Rightarrow n_{1} C_{v} (500 - 400) = 0.1 \times C_{v} (700 - 500)$
$\Rightarrow n_{1} \times 100 = 0.1 \times 200$
$n_{1} = 0.2$