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A vertical cylinder closed at both ends is fitted with a smooth piston dividing the volume into two parts each containing one mole of air. At the equilibrium temperature of 320 \mathrm{~K}, the upper and lower parts are in the ratio 4: 1. The ratio will become 3: 1 at a temperature of -

Option: 1

450K


Option: 2

228K


Option: 3

420K


Option: 4

570K


Answers (1)

\left(P_2-P_1\right) A=m g

or, \frac{m g}{A}=\frac{R T_f}{V_2}-\frac{R T_f}{3 r_2}=\frac{3 R T_f}{3 r_1}(second case) - (i)\\ \frac{m g}{A}=\frac{R T_i}{V_1}-\frac{R T_i}{4 V_1}=\frac{3 R T_i}{4 V_1}--(ii)

Further, 5 v_1=4 v_2

Equation both (i) and (ii), we get

\frac{3 T_i}{4 v_1}=\frac{2 T_f}{3 v_2}

\begin{aligned} \Rightarrow T_f=\frac{9}{8} \times \frac{V_2}{V_1} \times T_i & =\frac{9}{8} \times \frac{5}{4} \times 320 \\ & =450 \mathrm{k} \end{aligned}

Posted by

Kshitij

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