Gay-Laussac’s law

Gay-Laussac’s law (1809): according to this law, “when gases react, the volume of these gases, and the volumes of the product formed (if gaseous) are in simple whole number ratio to each other”.

\underset{1 vol.}{H_2(g)} + \underset{1 vol.}{Cl_2 (g)} \to \underset{2 vol.}{2HCl (g)}

The ratio 1 : 1 : 2 (a simple ratio)

Example 1. An iron cylinder contains helium at a pressure of 240 kPa at 300 K. The cylinder can withstand a pressure of  \times 10^6 Pa. The room in which cyliner is placed catches fire. Predict whether the cylinder will blow up before it melts or not? (M.P. of the cylinder I800 K)

Solution: According to pressure temperature relationship:

\dfrac{P_1}{T_1} = \dfrac{P_2}{T_2} \\[3mm] \dfrac{250 k Pa}{300K} = \dfrac{P_2}{1800 k} \\[3mm] \text{or} \hspace{20mm} P_2 = \dfrac{250 K Pa \times 1800 K}{300 K} \\[3mm] = 1.5 \times 10^6 Pa


But the cylinder will burst at 1.0 \times 10^6 Pa therefore the cylinder will burst before melting.


Example 2. From 0.2 g of CO_2 gas, 10^{21} molecules are removed. Calculate the number of moles ofCO_2 left behind.

Solution: Since gram molecular mass =1 mole

\because 44 g (mol. Mass) of CO_2 1 mole.

\therefore 0.2 g of CO_2 = \dfrac{1 mole \times 0.2 g}{44 g} \\ = 4.55 \times 10^{-3} mole


Since 6.023 \times 10^{23} molecules = 1 mole

1o^{21} molecules = \dfrac{1 \text{mole} \times 10^{21}}{6.023 \times 10^{23}} \\[3mm] = 1.66 \times 10^{-3} \text{mole} \\[3mm] \therefore \text{moles of} CO_2 \text{left} = 4.55 \times 10^{-3} \text{mole}- 1.66 \times 10^{-3} \text{mole} \\[3mm] 2.89 \time10^{-3} \text{mole}

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