Question

If a certain mass of mercury has a volume of 0.002 m3 at a temperature of 20°C, what will be the volume at 50°C?

A. 0.002010812m3
B. 0.004021624m3
C. 0.002021624m3
D. 0.000010812m3

Answer 1

The answer to this question lies in understanding the Combined Gas Law in chemistry, which combines Boyle's, Charles', and Gay-Lussac's gas laws. This law shows the relationship between pressure, volume, and temperature of gases and is represented by the equation P1V1/T1 = P2V2/T2. Changes in pressure (P) and volume (V) are directly proportional to one another, but these are inversely proportional to changes in temperature (T). This question must assume that pressure is constant, so the Ps cancel out and the equation becomes V1/T1 = V2/T2. Additionally, temperature should be converted to Kelvins (K) by adding 273 to the temperature in Celsius (so 20 C = 293 K, and 50 C = 323 K). If V1 = 0.002 and T1 = 293 and T2 = 323, then we solve the equation as V2 = T2*V1/T1 or V2 = (323*0.002)/293, so the answer is approximately equal to 0.0022... or answer (C).

Answer 2

As per the question the volume of mercury is given as 0.002 m^3 at 20 degree Celsius.

We are asked to calculate the volume of the mercury at 50 degree Celsius.

This problem is based on thermal expansion of matter.

Let us consider the initial and final volume of the mercury is denoted as -

`v_(1) \ and\ v_(2)`

Let the initial and final temperature of the mercury is denoted as -

`T_(1)\ and \ T_(2)`

As per question

`v_(1) =0.002 m^3`
`v_(2) =?`

`T_(1) =20^0 C`
`T_(2) =50^0 C`

The change in temperature is

`T_(2) -T_(1)`

`= 50^0 C -20^0 C`

`= 30^0 C`

Mercury is a fluid.So we have to apply volume expansion of liquid .

The coefficient of of volume expansion of mercury
`[ \gamma ]` at 20 degree Celsius is 0.00018 per centigrade.

As per volume expansion of liquid,

`V_(T) = v_(1) [1 +\gamma [T_(2) -T_(1) ]]`

Here
`V_(T)` is the volume at T degree Celsius.

Hence volume at 50 degree Celsius is calculated as-

`v_(2) =v_(1) [1+\gamma[50-30]]`

`= 0.002[1+0.00018*30]`

`=0.0020108m^3` [ans]

As per the options given in the question ,option A is close to the calculated value. So option A is right.