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We would like to make a vessel whose vo...

We would like to make a vessel whose volume does not change with temperature (take a hing from the problem aobve). We can use brass and iron `(beta_(brass) = 6 xx10^(-5) // K and beta_(iron) = 3.55 xx 10^(-5)//K)` ot create a volume of 100 c c. How do you think you can achieve this.

Text Solution

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Let `V__(io), V_(bo)` be the volume of iron and brass vessel at `0^@C.`
`V_i, V_b` be the volume of iron and brass vessel at Delta theta ` .^@C`,
`gamma_i, gamma_b` be the coeff. Of volume expansion of iron and brass. … (i)
As per question, `V_(io) - V_(bo) = 100 c.c = V_i - V_b`
Now `V_i = V_(io) (1 + gamma_i Delta theta)`
`V_b = V_(bo) (1 + gamma_b Delta theta)`
`V_i - V_b = (V_(io) - V_(bo)) + Delta theta (V_(io) gamma_i - V_(bo) gamma_(b))`
Since `V_i - V_b =` constant, so
`V_(io) gamma_i = V_(bo) gamma_b` or `(V_(io))/(V_(bo)) = (gamma_b)/(gamma_i) = ((3)/(2)beta_b)/((3)/(2)beta_i) = (beta_b)/(beta_i) = (6xx10^(-5))/(3.55 xx 10^(-5)) = (6)/(3.55)`
`:. (V_(io))/(V_(bo)) = (6)/(3.55)`..... (ii)
Solving (i) and (ii) we get
`V_(bo) = 144.9 c.c.` and `V_(io) = 244.9 c.c.`
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