Question

A certain metal has a coefficient of linear expansion of 2.00 × 10-5 K-1. It has been kept in a laboratory oven at 325°C for a long time. It is now removed from the oven and placed in a freezer at -145°C. After it has reached freezer temperature, the percent change in its density during this process is closest to

A) +2.90%.

B) -2.90%.

C) +2.74%.

D) -2.74%.

The anwser is A. I need an explanation

Answer 1

ろ d' d d o 木2-5-/-

Answer 2

if $\alpha$ is the coefficient of linear expansion ,$\beta$ is the coefficient of surface expansion and $\gamma$ is the coefficient of volumetric expansion then the relation between the $\alpha$ ,$\beta$ ,$\gamma$ is

  $\alpha$ /1 =$\beta$/2 = $\gamma$ /3 from this the relation between the $\alpha$ ,$\gamma$ is

  $\gamma$ = 3$\alpha$   -----------------------(1)

if V1 ,d1 are the intial volume and density and V2 ,d2 are the final volume and density then

from density (d) = mass (m) / volume (V)   

d V = constant then

d1 V1 =d2 V2

d2 /d1 = V1/V2 -----------------------(2)

from the volumetric compression the final volume V2 of the metal is obtained from below equation

V2 = V1 (1 - $\gamma$$\Delta$T) here $\gamma$ is replace with 3$\alpha$ from (1) we get

   V2 = V1 (1 - 3$\alpha$  $\Delta$T) and

V2 /V1 = 1 - 3$\alpha$  $\Delta$T   

reverse the above equation we get

V1 /V2 = 1 / (1 - 3$\alpha$  $\Delta$T )

using (2) replace  V1 /V2 by d2/d1 in above equation then

  d2/d1 = 1 / (1 - 3$\alpha$  $\Delta$T )

now subtract 1 from both sides of the equation we get

   (d2/d1) -1 = [1 / (1 - 3$\alpha$  $\Delta$T ) ] - 1

(d2-d1) / d1 =(3$\alpha$$\Delta$T) / (1 - 3$\alpha$  $\Delta$T ) ......... { d2-d1 =$\Delta$ d }

  $\Delta$ d /d = (3$\alpha$$\Delta$T) / (1 - 3$\alpha$  $\Delta$T )

here $\alpha$ =2.00 × 10^-5 K^-1   and $\Delta$T = (325 - (-145 )) K =470 K (difference will be same in both celsius and kelvin)

On substituting above values we get the percent of density increased with postive value ( Here d2> d1 , because during the compression of metal volume decreases and density increases)

  ($\Delta$ d /d ) x100 = (3 x2.00 × 10^-5 K^-1 x470K ) ( 1- ( 3 x2.00 × 10^-5 K^-1 ​x470K)) x100

  ($\Delta$ d /d ) x100 = + 2.90 %

Therefore the % of change in density is   + 2.90 %

so option (A) is the correct one .