Question

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

Answer 1

The formula of thermal volume expansion is given below:

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.

Answer 2

Answer:

Explanation:

T1=0°c=273K

T2=200°c=200+273=473K

Y (gama)=0.18×10^-3

D1=13.6

D=D1/1+y×delta T

Delta T=T2-T1=473-273=200

D=13.6/1+0.18×10^-3×200

D=13.6/1.036=13.127

●D=13.127●

◇REGARDS..........◇!!!!!