A certain quantity of real gas occupies a vol | vedclass

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(B) The compressibility factor $Z$ is defined as $Z = \frac{PV}{nRT}$,which implies $n = \frac{PV}{ZRT}$.Since the amount of gas $n$ remains constant,

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$393$

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(B) The compressibility factor $Z$ is defined as $Z = \frac{PV}{nRT}$,which implies $n = \frac{PV}{ZRT}$.Since the amount of gas $n$ remains constant,we have $\frac{P_1 V_1}{Z_1 T_1} = \frac{P_2 V_2}{Z_2 T_2}$.Given values: $P_1 = 100 \ atm$,$V_1 = 0.15 \ dm^3$,$T_1 = 500 \ K$,$Z_1 = 1.07$.$P_2 = 300 \ atm$,$T_2 = 300 \ K$,$Z_2 = 1.4$.Substituting these values into the equation:$\frac{100 	imes 0.15}{1.07 	imes 500} = \frac{300 	imes V_2}{1.4 	imes 300}$$V_2 = \frac{100 	imes 0.15 	imes 1.4 	imes 300}{1.07 	imes 500 	imes 300}$$V_2 = \frac{15 	imes 1.4}{1.07 	imes 500} = \frac{21}{535} \approx 0.03925 \ dm^3$.Converting to $10^{-4} \ dm^3$ units: $0.03925 	imes 10^4 	imes 10^{-4} = 392.5 	imes 10^{-4} \ dm^3$.Rounding to the nearest integer,we get $393 	imes 10^{-4} \ dm^3$.

$A$ certain quantity of real gas occupies a volume of $0.15 \ dm^3$ at $100 \ atm$ and $500 \ K$ when its compressibility factor is $1.07$. Its volume at $300 \ atm$ and $300 \ K$ (when its compressibility factor is $1.4$) is $........ \times 10^{-4} \ dm^3$ (Nearest integer).

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