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JEE Advanced 2011Paper 1
Exam
JEE Advanced
Year
2011
Paper
Paper 1
Subject
Chemistry
Chapter
States of Matter

JEE Advanced 2011 · Paper 1 · Chemistry · States of Matter

To an evacuated vessel with movable piston under external pressure...

To an evacuated vessel with movable piston under external pressure of 1 atm1 \mathrm{~atm}, 0.10.1 mole of He and 1.01.0 mole of an unknown compound (vapour pressure 0.68mathrm atm0.68 mathrm{~atm} at 0∘C0^{\circ} \mathrm{C} ) are introduced. Considering the ideal gas behaviour, the total volume (in litre) of the gases at 0∘C0^{\circ} \mathrm{C} is close to

Correct answer

7

Step-by-step solution

Since, the external pressure is 1.0 atm1.0 \mathrm{~atm}, the gas pressure is also 1.0 atm1.0 \mathrm{~atm} as piston is movable. Out of this 1.0 atm1.0 \mathrm{~atm} partial pressure due to unknown compound is 0.68 atm0.68 \mathrm{~atm}. Therefore, partial pressure of He=1.00−0.68=0.32 atm\mathrm{He}=1.00-0.68=0.32 \mathrm{~atm}. ⇒ Volume =n(He)RTr(He)=0.1×0.082×2730.32=7 L⇒ Volume of container = Volume of He.\begin{aligned}& \Rightarrow \quad \text{ Volume }=\frac{n(\mathrm{He}) \mathrm{RT}}{r(\mathrm{He})}=\frac{0.1 \times 0.082 \times 273}{0.32}=7 \mathrm{~L}\\ & \Rightarrow \text{ Volume of container }=\text{ Volume of } \mathrm{He} . \end{aligned}
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