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!!!

The density increases by 5 times

Explanation:

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

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

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:

So, the volume decreases by 5 times.

The density of the gas is given by

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.

if the concentration of products is decreased, the equilibrium will shift toward the products). add an inert gas (one that is not involved in the reaction) to the constant-volume reaction mixture: this will increase the total pressure of the system, but will have no effect on the equilibrium condition.

explanation:

your answer is a. the enthalpy of formation for a pure element in its standard state is always positive.