(B) Using the ideal gas equation $PV = nRT$. Given: $n = 1 \ mol$, $V = 12 \ L = 12 \times 10^{-3} \ m^3$, $T = 297 + 273 = 570 \ K$, $R = 8.314 \ J \

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(B) Using the ideal gas equation $PV = nRT$. Given: $n = 1 \ mol$, $V = 12 \ L = 12 \times 10^{-3} \ m^3$, $T = 297 + 273 = 570 \ K$, $R = 8.314 \ J \ K^{-1} \ mol^{-1}$. $P = \frac{nRT}{V} = \frac{1 \times 8.314 \times 570}{12 \times 10^{-3}} \ Pa$. $P = \frac{4738.98}{0.012} \ Pa = 394915 \ Pa \approx 395 \ kPa$.

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

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One mole of an ideal gas occupies $12 \ L$ at $297^{\circ} C$. What is the pressure of the gas?

TS EAMCET 2020 All TS EAMCET

A
$207 \ kPa$
B
$395 \ kPa$
C
$395 \ Pa$
D
$207 \ Pa$

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States of Matter

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Ideal gas equation and Related gas laws

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One mole of an ideal gas occupies $12 \ L$ at $297^{\circ} C$. What is the pressure of the gas?

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If $500 \ cm^3$ of hydrogen gas is at a pressure of $760 \ mm$ and a temperature of $300 \ K$,what is the number of moles?

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