A steel ball that is initially at a pressure of 1.0 × 10^{5 }Pa is heated from 20°C to 120°C, keeping its volume constant.

Find the pressure inside the ball. Coefficient of linear expansion of steel = 12 × 10^{–6} °C^{–1}and bulk modulus of steel = 1.6 × 10^{11} Nm^{–2}.

#### Solution

Given:

Initial pressure on the steel ball = 1.0 × 10^{5 }Pa

The ball is heated from 20 °C to 120 °C.

So, change in temperature, Δθ = 100°C.

Coefficient of linear expansion of steel, α = 12 ×10^{-6}°C^{-1}

Bulk modulus of steel ,*B* = 1.6 × 10^{11} Nm^{–2}

Pressure is given as,

⇒ P = B × γ Δθ

⇒ P =B × 3 αΔθ (∵ γ = 3 α )

⇒ P = 1.6 × 10^{11} ×3 × 12 × 10^{-6} ×(120-20)

=1.6 × 3 × 12 ×10^{11} ×10^{-6} ×10^{2}

= 57.6 × 10^{7}

⇒ P =5.8 × 10^{8} Pa

Therefore, the pressure inside the ball is 5.8 × 10^{8 }Pa.