Question

State and explain Gay Lussac’s law.

Answer 1

It states that if the volume remains constant, the pressure of a given mass of a gas increases or decreases by its pressure at 0°C for each 1°C rise or fall in temperature.

If P₀ and P_t are the pressure of a given mass of gas at 0°C and t°C respectively, then according to Gay Lussac’s law,

P_t = P₀ (1+\(\frac{t}{273.15}\))

or P_t = P₀\((\frac{273.15+t}{273.15})\)

or P_t = P₀\(\frac{T}{T_0}\)

where T₀(K) = 273.15

and T(K) = 273.15 + t

\(\frac{P_t}{P_0}\) = \(\frac{T}{T_0}\)

\(\frac{P}{T}\) = constant

or P ∝ T