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A container A contains an ideal gas at pressure P' volume ' $\mathrm{V}$ ' and temperature ' $\mathrm{T}$ '. A second container B contains the same gas at pressure ' $2 \mathrm{P}$ ', volume ' $2 \mathrm{~V}$ ' and temperature $\frac{\mathrm{T}}{2}$. The ratio of mass of gas in $\mathrm{A}$ to that in $\mathrm{B}$ is
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The correct answer is:
$1: 8$
Ideal gas equation is:
$\mathrm{PM}=\rho \mathrm{RT}$
$\rho=\left(\frac{\mathrm{m}}{\mathrm{v}}\right)=\left(\frac{\text { mass }}{\text { volume }}\right), \mathrm{M}$ is molecular weight.
Taking ratio of $\mathrm{eq}^{\mathrm{n}}(1) \&(2)$
$\frac{\mathrm{m}_{\mathrm{A}}}{\mathrm{m}_{\mathrm{B}}}=\frac{1}{8}$
$\mathrm{PM}=\rho \mathrm{RT}$
$\rho=\left(\frac{\mathrm{m}}{\mathrm{v}}\right)=\left(\frac{\text { mass }}{\text { volume }}\right), \mathrm{M}$ is molecular weight.
Taking ratio of $\mathrm{eq}^{\mathrm{n}}(1) \&(2)$
$\frac{\mathrm{m}_{\mathrm{A}}}{\mathrm{m}_{\mathrm{B}}}=\frac{1}{8}$
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