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Standard VIII

Chemistry

Gases

The equation ...

Question

The equation of state for a gas is $P (V - n b) = n R$ , where $b$ and $R$ are constants. If the pressure and temperature are such that $V_{m} = 10 b$ , then what is the compressibility factor of gas in this condition ?

\frac{10}{9}

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\frac{9}{11}

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\frac{11}{10}

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\frac{10}{11}

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Solution

The correct option is A $\frac{10}{9}$
Given,
$V_{m} = 10 b$
We know that,
$P (V - n b) = n R T$
Compressibility factor $Z$ is given as:
$Z = \frac{P V_{m}}{R T} = \frac{V_{m}}{V_{m} - b} - \frac{a}{V_{m} R T}$
But $a = 0$
Hence,
$Z = \frac{P V_{m}}{R T} = \frac{V_{m}}{V_{m} - b} = \frac{10 b}{10 b - b} = \frac{10}{9}$

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The equation of state for a gas is P(V−nb)=nR, where b and R are constants. If the pressure and temperature are such that Vm=10b, then what is the compressibility factor of gas in this condition ?

The correct option is A 109Given, Vm=10b We know that, P(V−nb)=nRT Compressibility factor Z is given as: Z=PVmRT=VmVm−b−aVmRT But a=0 Hence, Z=PVmRT=VmVm−b=10b10b−b=109

The equation of state for a gas is $P (V - n b) = n R$ , where $b$ and $R$ are constants. If the pressure and temperature are such that $V_{m} = 10 b$ , then what is the compressibility factor of gas in this condition ?