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 $Δ$ $∆$ V(<<V)of air is transferred to the tube adiabatically. What is the work done when the pressure in the tube is increased from $p_{1} t o p_{2}$ $p_{1} t o p_{2}$ ?

Hint: Use the equation of the adiabatic process.

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

Let, the volume is increased by

Δ

$∆$ V and pressure is increased by

Δ

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

p_{1} {V_{1}}_{1}^{γ} = p_{2} {V_{2}}_{2}^{γ}

$p_{1} {V_{1}}^{γ} = p_{2} {V_{2}}^{γ}$

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

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

∵

$∵$ volume is fixed)

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

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

(1 + \frac{Δ p}{p}) V^{γ}

$(1 + \frac{∆ p}{p}) V^{γ}$

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

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

∵ Δ

$∵ ∆$ v<<v)

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

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

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

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

Step 2: Find the work done.

Hence, the work done in increasing the pressure from

p_{1} t o p_{2}

$p_{1} t o p_{2}$ :

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

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

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

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

\Rightarrow

$\Rightarrow$

W = \frac{(p_{2} - p_{1})}{γ} V

$W = \frac{(p_{2} - p_{1})}{γ} V$

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