A) 50 mL
B) 100 mL
C) 200 mL
D) 400 mL
Correct Answer: B
Solution :
Given,\[{{p}_{1}}=100\text{ }mm,\text{ }{{V}_{1}}=200\text{ }mL\text{ }and\text{ }{{p}_{2}}=400\]mm From Boyle' Law \[{{p}_{1}}{{V}_{1}}={{p}_{2}}{{V}_{2}}\] \[{{V}_{2}}=\frac{{{p}_{1}}{{V}_{1}}}{{{p}_{2}}}\] \[=\frac{100\times 200}{400}\] \[{{V}_{2}}=50\,mL\] Volume of 2 mole gas \[=2\times 50\] \[=100mL\]
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