Question

What would happen to the density of an ideal gas if its pressure is cut in half and its Kelvin temperature decreases by 10 times the original temperature? Be sure to explain in words and numbers!!!

Answer 1

Answer:

The density increases by 5 times

Explanation:

We can solve this problem by using the equation of state of an ideal gas:

$pV=nRT$

where

p is the pressure of the gas

V is its volume

n is its number of moles

R is the gas constant

T is the Kelvin temperature

Since n and R are constant during a gas transformation, we can rewrite the equation as

$\frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2}$

In thhis problem we have:

$p_2=\frac{p_1}{2}$, since the pressure of the gas is cut in half

$T_2=\frac{T_1}{10}$, since the gas temperature decreases by 10 times

Therefore solving for V2, we find how much does the volume of the gas change:

$V_2=\frac{p_1 V_1 T_2}{p_2 T_1}=\frac{p_1 V_1 (T_1/10)}{(p_1/2) T_1}=\frac{V_1}{5}$

So, the volume decreases by 5 times.

The density of the gas is given by

$d=\frac{m}{V}$

where m is the mass of the gas and V its volume. Here, the mass of the gas remains constant: so, the density is inversely proportional to the volume. Therefore, if the volume decreases by 5 times, the density will increase by 5 times.