JEE Advanced · Chemistry · 5. States of Matter
To an evacuated vessel with movable piston under external pressure of \(1 \mathrm{~atm}\), \(0.1\) mole of He and \(1.0\) mole of an unknown compound (vapour pressure \(0.68 mathrm{~atm}\) at \(0^{\circ} \mathrm{C}\) ) are introduced. Considering the ideal gas behaviour, the total volume (in litre) of the gases at \(0^{\circ} \mathrm{C}\) is close to
- A 1
- B 3
- C 5
- D 7
Answer & Solution
Correct Answer
(D) 7
Step-by-step Solution
Detailed explanation
Since, the external pressure is \(1.0 \mathrm{~atm}\), the gas pressure is also \(1.0 \mathrm{~atm}\) as piston is movable. Out of this \(1.0 \mathrm{~atm}\) partial pressure due to unknown compound is \(0.68 \mathrm{~atm}\). Therefore, partial pressure of \(\mathrm{He}=1.00-0.68=0.32 \mathrm{~atm}\).
\(\Rightarrow \text {Volume }=\frac{n(\mathrm{He}) \mathrm{RT}}{r(\mathrm{He})}=\frac{0.1 \times 0.082 \times 273}{0.32}=7 \mathrm{~L} \)
\( \Rightarrow \text {Volume of container }=\text { Volume of } \mathrm{He}.\)
\(\Rightarrow \text {Volume }=\frac{n(\mathrm{He}) \mathrm{RT}}{r(\mathrm{He})}=\frac{0.1 \times 0.082 \times 273}{0.32}=7 \mathrm{~L} \)
\( \Rightarrow \text {Volume of container }=\text { Volume of } \mathrm{He}.\)
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