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](https://tex.z-dn.net/?f=pV%3DnRT)
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}](https://tex.z-dn.net/?f=%5Cfrac%7Bp_1%20V_1%7D%7BT_1%7D%3D%5Cfrac%7Bp_2%20V_2%7D%7BT_2%7D)
In thhis problem we have:
, since the pressure of the gas is cut in half
, 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}](https://tex.z-dn.net/?f=V_2%3D%5Cfrac%7Bp_1%20V_1%20T_2%7D%7Bp_2%20T_1%7D%3D%5Cfrac%7Bp_1%20V_1%20%28T_1%2F10%29%7D%7B%28p_1%2F2%29%20T_1%7D%3D%5Cfrac%7BV_1%7D%7B5%7D)
So, the volume decreases by 5 times.
The density of the gas is given by
![d=\frac{m}{V}](https://tex.z-dn.net/?f=d%3D%5Cfrac%7Bm%7D%7BV%7D)
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.