Question

Suppose that you lowered the temperature of a gas from 100˚C to 50 ˚C. By what factor do you change the volume of the gas?

Answer 1

Answer:- Volume decreases by a factor of 1.15.

Solution:- At constant pressure, volume of the gas is directly proportional to the the kelvin temperature.

The equation is written as:

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

where, $V_1$ is the volume at initial temperature $T_1$ and $V_2$ is the volume at final temperature $T_2$ .

Temperature must be in kelvins. So, let's convert both the temperatures to kelvin.

To convert degree C to kelvin we add 273.

So, $T_1$  = 100 + 273 = 373 K

$T_2$ = 50 + 273 = 323 K

The equation could also be written as:-

$\frac{V_1}{V_2}=\frac{T_1}{T_2}$

$\frac{V_1}{V_2}=\frac{373}{323}$

$\frac{V_1}{V_2}$ = 1.15

From here we could say that the volume decreases by a factor of 1.15.

For example if the initial volume $V_1$ is 1 L then final volume $V_2$ will be $\frac{1L}{1.15}$ that is 0.87 L.