Ncert Exercises & Solutions

Consider a cycle tyre being filled with air by a pump. Let V be the volume of the tyre (fixed) and at each stroke of the pump $∆$ V(<<V)of air is transferred to the tube adiabatically. What is the work done when the pressure in the tube is increased from $p_{1} t o p_{2}$ ?

Hint: Use the equation of the adiabatic process.

Step 1: Find the change in volume of the tyre.

Let, the volume is increased by

∆

V and pressure is increased by

∆

p by a stroke. For just before and after a stroke, we can write,

p_{1} {V_{1}}^{γ} = p_{2} {V_{2}}^{γ}

\Rightarrow p {(V + ∆ V)}^{γ} = (p + ∆ p) V^{γ}

(

∵

volume is fixed)

\Rightarrow p V^{γ} {(1 + \frac{∆ V}{V})}^{γ} = p

(1 + \frac{∆ p}{p}) V^{γ}

\Rightarrow p V^{γ} (1 + γ \frac{∆ V}{V}) = p V^{γ} (1 + \frac{∆ p}{ρ})

(

∵ ∆

v<<v)

\Rightarrow γ \frac{∆ V}{V} = \frac{∆ p}{p} \Rightarrow ∆ V = \frac{1}{γ} \frac{V}{p} ∆ p

\Rightarrow d V = \frac{1}{γ} \frac{V}{p} d p

Step 2: Find the work done.

Hence, the work done in increasing the pressure from

p_{1} t o p_{2}

W = \int_{p_{1}}^{p_{2}} p d V = \int_{p_{1}}^{p_{2}} p \times \frac{1}{γ} \frac{V}{p} d p

= \frac{V}{γ} \int_{p_{1}}^{p_{2}} d p = \frac{V}{γ} (p_{2} - p_{1})

\Rightarrow

W = \frac{(p_{2} - p_{1})}{γ} V

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