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We would like to make a vessel whose volume does not change with temperature. We can use brass and iron $\left (\beta _{brass} = \frac{6 \times 10^{-6}}{K} and \beta _{iron} = \frac{3.55 \times 10^{-5}}{K} \right )$ to create a volume of 100 cc. How do you think you can achieve this?

Answers (1)

For, the above situation, we need to make a double container whose volume difference will be 100cc.

Let us consider $V_{1i}$ & $V_{1b}$ as the initial volumes of the iron and brass container, i.e., $V_{1i}- V_{1b} =100cc$ .

After heating by $\Delta TK$ the difference will be the same, but the new volumes will be V2i &V2b.

Now, $V_{2i}- V_{2b} =100cc$

$\gamma = \frac{\Delta V}{V\Delta T}$

Therefore, $V_{2} - V_{1} = \gamma V_{1}\Delta T$

$V_{2} =V_{1}+ \gamma V_{1}\Delta T$

$= V_{1} (1 + \gamma \Delta T)$

$V_{2i}= V_{1i} (1 + \gamma_{i} \Delta T)$

& $V_{2b} = V_{1b} (1 + \gamma _{b}\Delta T)$

$V_{1i} + V_{1i}\gamma _{i}\Delta T - (V_{ib} + V_{ib}\gamma _{b}\Delta T) = 100 cc.$

$100 + (V_{1i}\gamma _{i} - V_{1b}\gamma _{b}) \Delta T = 100$

$V_{1i}\gamma _{i} - V_{1b}\gamma _{b} = 0$

$\frac{V_{1i}}{V_{1b} }= \frac{\gamma _{b}}{\gamma _{i}}$

$=\frac{ 6 \times 10^{-5}}{3.55 \times 10^{-5}}$

$=\frac{ 6 }{3.55 }$

$=\frac{ 120}{71 }$

Now, let $V_{1i} = 120x , V_{1b} = 71x$

& $V_{1i} - V_{1b} = 100$

Thus, 120x – 71x = 100

49x = 100

$x =\frac{100}{49}$

= 2.04

Thus, $V_{1i} = 120 \times 2.04 = 245 cc$

$V_{1b} = 71 \times 2.04 = 145 cc$

Posted by

infoexpert24

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