Study of Gas Laws — Question 2

(a) Charles' law — It states that pressure remaining constant, the volume of a given mass of dry gas increases or decreases by $\dfrac{1}{273}$ of its volume at 0°C for each 1°C increase or decrease in temperature, respectively.

(b) Let V₀ be the volume of a fixed mass of a gas at 0°C and let V be its volume at temperature t°C at constant pressure. Then according to Charles' law,

V = V_o + $\dfrac{\text{V}_0}{273}$t (when P is constant)

V = V_o (1 + $\dfrac{\text{t}}{273}$)

= V_o ($\dfrac{\text{273 + t}}{273}$)

V = $\dfrac{\text{V}_0}{273}$T where T = 273 + t

For a given mass of a gas,

$\dfrac{\text{V}_0}{273}$ = constant

∴ V = k x T (where k is constant)

or V α T and $\dfrac{\text{V}}{\text{T}}$ = K

Suppose a gas occupies V₁ cm³ at T₁ temperature and V₂ cm³ at T₂ temperature, then by Charles law,

    V₁ α T₁

or V₁ = kT₁ (k is constant)

or $\dfrac{\text{V}_1}{\text{T}_1}$ = k and

    V₂ α T₂

or $\dfrac{\text{V}_2}{\text{T}_2}$ = k

∴ $\dfrac{\text{V}_1}{\text{T}_1}$ = $\dfrac{\text{V}_2}{\text{T}_2}$ = k (at constant pressure)

(c) Volume of a given mass of a gas is directly proportional to its temperature, hence, density decreases with an increase in temperature. This is the reason that hot air is filled into balloons used for meteorological purposes.