Physics, asked by sapnasofty9368, 1 year ago

Coefficient+of+volume+expansion+of+mercury+is+0.18×10^-3/℃+if+the+density+of+mercury+at+0℃+is+13.6+g/cc+then+its+density+at+200℃+is

Answers

Answered by loveguru38
4

Answer:

Here your answers

v= v_0*(1+y*dt) where v_0 is initial volume, y is coefficient of volume expansion of mercury, and dt is change in temperature.

Putting the values of known variables, we get

v= v_0*(1+0.18x10^(-3)*(200-0))

v= v_0*(1+3.6x10^(-2))

v= v_0*(1+0.036)

v= v_0*(1.036)

Now, divide the above equation by mass of mercury, m

v/m= (v_0/m)*(1.036)

We know, m/v represents density.

So,

1/d= (1/d_0)*(1.036) where m/v= d and m/v_0= d_0

Rearranging the above equation, we get

d= d_0/1.036

Given that the density of mercury at 0℃ is 13.6 g/cc

So,

d= 13.6/1.036

d= 13.127 g/cc

Therefore, the density of mercury at 200℃ would become 13.127 g/cc.

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