Question

If 22 L of helium at 20.0°C is allowed to expand to 44 L, with the pressure remaining the same, what is the new temperature in Celsius?

Answer 1

Using Charle's Law,

V₁ = 22L

T₁ = 20°C + 273 = 293K

V₂ = 44L

T₂ = ?

$\dfrac{V_1}{T_1} = \dfrac{V_2}{T_2}$

$\dfrac{22L}{293K} = \dfrac{44L}{T_2}$

$T_2 = \dfrac{293×44}{22} = 586K$

T₂ (°C) = 586K -273 = 313°C