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Standard IX

Chemistry

Gaseous State

Helium gas at...

Question

Helium gas at 300 K is shifted from a vessel of $250 c m^{3}$ to a vessel of $1 L$ capacity. The pressure of gas will

become four times

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become double

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decrease to half of the original value

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decrease to one-fourth the original value

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Solution

The correct option is D
decrease to one-fourth the original value
$P V = C o n s t a n t Boyle's Law : P_{1} V_{1} = P_{2} V_{2} V_{1} = 250 c m^{3} \Rightarrow \frac{1}{4} L V_{2} = 1 L P_{1} = P P_{2} = ? P \times \frac{1}{4} = P_{2} \times 1 P_{2} = \frac{P}{4}$

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Helium gas at 300 K is shifted from a vessel of 250 cm3 to a vessel of 1 L capacity. The pressure of gas will

The correct option is D decrease to one-fourth the original valuePV=ConstantBoyle's Law:P1V1=P2V2V1=250 cm3⇒14LV2=1LP1=PP2=?P×14=P2×1P2=P4

Helium gas at 300 K is shifted from a vessel of $250 c m^{3}$ to a vessel of $1 L$ capacity. The pressure of gas will

The correct option is D
decrease to one-fourth the original value
$P V = C o n s t a n t Boyle's Law : P_{1} V_{1} = P_{2} V_{2} V_{1} = 250 c m^{3} \Rightarrow \frac{1}{4} L V_{2} = 1 L P_{1} = P P_{2} = ? P \times \frac{1}{4} = P_{2} \times 1 P_{2} = \frac{P}{4}$