Question

If a cylinder have 640cm3 of gas in it at 0°C, then how many parts of its volume increases or decreases by 1°C rise or fall in temperature?​

Answer 1

Explanation:

This point, with a temperature of –273.15 ˚C, is the theoretical point where the samples would have “zero volume”. This temperature, -273.15 ˚C, is called absolute zero.

Answer 2

The temperature should be raised by $1^{0}C$ to soluble the volume.

Explanation:

$10^{0} C = 640 K (because 0^{0}C = 3 K)$

Now to solve this problem, you can use combined gas equation.

 $\frac{P1V1}{T1} =\frac{P2V2}{T2}$

Here, there is no discussion about pressure so we can assume that the pressure is constant. So $P1 = P2 = P.$

 $\frac{PV1}{T1} \frac{PV2}{T2}$ $\frac{V1T1}{V2T2}$ … … … (1)

Now, $V2 = 2V1 and T1 = 640K$

Putting these values in (1) we get

$\frac{V1}{640} =\frac{2V1}{T2}$

 $T2 = 2 × 640=1280 K = 1^{0} C$

 If you put the value of T in ℃ you will get wrong answer. In problems related to thermodynamics always put the value of T in Kelvin only.