Question

A large, 50:0 cm diameter metal cylinder filled with air supports a 20:0 Kg piston that can slide up and down without friction. The piston is 100.0 cm above the bottom when the temperature is 200º C. An 80.0 Kg student then stands on the piston. After several minutes have elapsed, by how much has the piston been depressed?

Answer 1

Answer:

The piston has been depressed by 80 cm

Explanation:

GIVEN

• Diameter of cylinder D = 50 cm
• Piston of 20 Kg is supported by the air inside the cylinder when height h₁ = 100 cm
• Mass of student on piston, = 80Kg

CALCULATION

Now

$P₁ = \frac{20g}{A}$

$V₁ = Ah₁ = 100A$

$P₂ = \frac{(20+80)g}{A}$

$P₂ = \frac{100g}{A}$

Let us consider that the air undergoes isothermal process when the extra weight is placed on the piston.From Ideal Gas Equation

$PV = nRT = constant$

$or\:\:P₁V₁ = P₂V₂$

$or\:\:\frac{20g}{A}\times100A= \frac{100g}{A} Ah₂$

$or\:\:20\times100 = 100h₂$

$or\:\:h₂ = 20\:\:cm$

Hence the piston has been depressed by

$Δh = h₁ - h₂$

$Δh = 100 - 20$

$Δh = 80\:\:cm$