(B) According to the ideal gas law,$\frac{p_{1} V_{1}}{T_{1}} = \frac{p_{2} V_{2}}{T_{2}}$.Given: $T_{2} = 2 T_{1}$ and $p_{2} = \frac{p_{1}}{2}$.Subs

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will be $4$ times the original volume

(B) According to the ideal gas law,$\frac{p_{1} V_{1}}{T_{1}} = \frac{p_{2} V_{2}}{T_{2}}$.Given: $T_{2} = 2 T_{1}$ and $p_{2} = \frac{p_{1}}{2}$.Substituting these values into the equation:$\frac{p_{1} V_{1}}{T_{1}} = \frac{(p_{1} / 2) \times V_{2}}{2 T_{1}}$.Canceling $p_{1}$ and $T_{1}$ from both sides:$V_{1} = \frac{V_{2}}{4}$.Therefore,$V_{2} = 4 V_{1}$.The volume becomes $4$ times the original volume.

Class 11 Chemistry States of Matter Ideal gas equation and Related gas laws

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When the absolute temperature of an ideal gas is doubled and the pressure is halved,the volume of the gas

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A
will be half of the original volume
B
will be $4$ times the original volume
C
will be $2$ times the original volume
D
will be $1/4$ th of the original volume

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Ideal gas equation and Related gas laws

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When the absolute temperature of an ideal gas is doubled and the pressure is halved,the volume of the gas

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An ideal gas expands isothermally and reversibly from $10 \ m^3$ to $20 \ m^3$ at $300 \ K$,performing $5.187 \ kJ$ of work on the surroundings. Calculate the number of moles of gas used.

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An ideal gas on isothermal reversible compression from $10 \ L$ to $5 \ L$ performs $1730 \ J$ of work at $300 \ K$. Calculate the number of moles of gas involved in the compression? $(R = 8.314 \ J \ K^{-1} \ mol^{-1})$

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