PYQ NEET- வேதியியல் சமநிலை-1 L-2

Question: $ A 20 litre container at $400 \mathrm{~K}$ contains $\mathrm{CO}_2(g)$ at pressure $0.4 \mathrm{~atm}$ and an excess of $\mathrm{SrO}$ (neglect the volume of solid $\mathrm{SrO}$ ). The volume of the container is now decreased by moving the movable piston fitted in the container. The maximum volume of the container, when pressure of $\mathrm{CO}_2$ attains its maximum value, will be#

(Given that :
$$ \begin{array}{r} \mathrm{SrCO}_3(\mathrm{~s}) \rightleftharpoons \mathrm{SrO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g})_1 \ \left.\mathrm{~K}_p=1.6 \mathrm{~atm}\right) \end{array} $$

A) $5 \mathrm{~L}$

B) $10 \mathrm{~L}$

C) $4 \mathrm{~L}$

D) $2 \mathrm{~L}$

Answer: $5 \mathrm{~L}$#

Solution:#

For the reaction,
$$ \begin{aligned} & \mathrm{SrCO}_3(s) \rightleftharpoons \mathrm{SrO}(s)+\mathrm{CO}_2(q), \ & \mathrm{K}p=1.6 \mathrm{~atm}=p{\mathrm{CO}_2}=\text { maximum } \end{aligned} $$
pressure of $\mathrm{CO}_2$
Given,
$$ \begin{aligned} & p_1=0.4 \mathrm{~atm}, V_1=20 \mathrm{~L}, T_1=400 \mathrm{~K} \ & p_2=1.6 \mathrm{~atm}, V_2=?, T_2=400 \mathrm{~K} \end{aligned} $$

At constant temperature, $p_1 V_1=p_2 V_2$
$$ \begin{aligned} 0.4 \times 20 & =1.6 \times V_2 \ V_2 & =\frac{0.4 \times 20}{1.6}=5 \mathrm{~L} \end{aligned} $$