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
if is the coefficient of linear expansion , is the coefficient of surface expansion and is the coefficient of volumetric expansion then the relation between the , , is
/1 =/2 = /3 from this the relation between the , is
= 3 -----------------------(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 - T) here is replace with 3 from (1) we get
V2 = V1 (1 - 3 T) and
V2 /V1 = 1 - 3 T
reverse the above equation we get
V1 /V2 = 1 / (1 - 3 T )
using (2) replace V1 /V2 by d2/d1 in above equation then
d2/d1 = 1 / (1 - 3 T )
now subtract 1 from both sides of the equation we get
(d2/d1) -1 = [1 / (1 - 3 T ) ] - 1
(d2-d1) / d1 =(3T) / (1 - 3 T ) ......... { d2-d1 = d }
d /d = (3T) / (1 - 3 T )
here =2.00 × 10-5 K-1 and 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)
( d /d ) x100 = (3 x2.00 × 10-5 K-1 x470K ) ( 1- ( 3 x2.00 × 10-5 K-1 x470K)) x100
( d /d ) x100 = + 2.90 %
Therefore the % of change in density is + 2.90 %
so option (A) is the correct one .
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