Question

We would like to make a vessel whose volume does not change with temperature (take a hint from the problem above). We can use brass and iron `(β_(vbrass) = (6 xx 10^(–5))/K and β_(viron) = (3.55 xx 10^(–5))/K)` to create a volume of 100 cc. How do you think you can achieve this.

Answer 1

In the previous problem, the difference in the length was constant.

In this problem the difference in volume is constant.

The situation is shown in the diagram.

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 Δθ°C,

γ_i, γ_b be the coefficient of volume expansion of iron and brass.

As per the question, V_io – V_bo = 100 cc = V_i – V_b  ......(i)

Now, `V_i = V_(io) (1 + γ_iΔθ)`

`V_b = V_(bo) (1 + γ_bΔθ)`

`V_i - V_b = (V_(io) - V_(bo)) + Δθ(V_(io)γ_i - V_(bo)γ_b)`

Since, `V_i - V_b` = constant

So, `V_(io)γ_i - V_(bo)γ_b`

⇒ `V_(io)/V_(bo) = γ_b/γ_i`

= `(3/2 β_b)/(3/2 β_i)`

= `β_b/β_i`

= `(6 xx 10^-5)/(3.55 xx 10^-5)`

= `6/3.55`

`V_(io)/V_(bo) = 6/3.55`  ......(ii)

Solving equations (i) and (ii), we get

V_io = 244.9 cc

V_bo = 144.9 cc