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NCERT Exemplar · Q20

Q.A rigid, thermally isolated container is divided into two chambers by a partition. The left chamber has volume V1=2.0V_1 = 2.0 litre and contains μ1=4.0\mu_1 = 4.0 moles of a gas at pressure p1=1.00p_1 = 1.00 atm; the right chamber has volume V2=3.0V_2 = 3.0 litre and contains μ2=5.0\mu_2 = 5.0 moles of the same gas at pressure p2=2.00p_2 = 2.00 atm. The partition is removed and the gas is allowed to mix and reach equilibrium. Because the container is isolated, the total internal energy of the gas is conserved. Calculate the pressure of the mixture after equilibrium is reached.

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Since the container is isolated, the gas's total internal energy is unchanged when the partition is pulled out. Writing each chamber's energy through μiTi=piVi/R\mu_iT_i = p_iV_i/R and equating the total before and after mixing gives P(V1+V2)=p1V1+p2V2P(V_1+V_2) = p_1V_1 + p_2V_2, so P=(2.0+6.0)/5.0=1.6P = (2.0 + 6.0)/5.0 = 1.6 atm.

Concept

For an ideal gas the internal energy depends only on temperature: U=μ CV TU = \mu\,C_V\,T. When the two gas samples mix in an isolated rigid box, no work is done on the surroundings and no heat leaves, so the total internal energy is conserved.

Why this approach

Let the initial temperatures be T1T_1 and T2T_2 and the final common temperature TT. From the ideal-gas law in each chamber:

p1V1=μ1RT1⇒μ1T1=p1V1R,p_1V_1 = \mu_1 R T_1 \quad\Rightarrow\quad \mu_1 T_1 = \frac{p_1V_1}{R},

p2V2=μ2RT2⇒μ2T2=p2V2R.p_2V_2 = \mu_2 R T_2 \quad\Rightarrow\quad \mu_2 T_2 = \frac{p_2V_2}{R}.

Energy conservation (CVC_V is the same for the same gas) gives:

μ1CVT1+μ2CVT2=(μ1+μ2)CVT,\mu_1 C_V T_1 + \mu_2 C_V T_2 = (\mu_1+\mu_2) C_V T,

⇒  (μ1+μ2) T=μ1T1+μ2T2=p1V1+p2V2R.\Rightarrow\; (\mu_1+\mu_2)\,T = \mu_1 T_1 + \mu_2 T_2 = \frac{p_1V_1 + p_2V_2}{R}.

Steps

  1. Apply the ideal-gas law to the whole box after mixing (total moles μ1+μ2\mu_1+\mu_2, total volume V1+V2V_1+V_2): P(V1+V2)=(μ1+μ2)RT.P(V_1+V_2) = (\mu_1+\mu_2) R T. …

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