need help understand how to do this chemistry problem?

Question

a gas at a pressure of 2.0 atm is contained in a closed container. Indicate the changes in its volume when the pressure increases to 6.0 atm.

Answer 1

Think about the ideal gas law. It's really helpful in solving problems like these.

PV = nRT

So, pressure and volume are proportional to the variables on the other side of the equation (which remain constant). If pressure increases, volume will have to DECREASE to equal the other side of the equation.

How much will volume decrease by? Because the pressure has tripled, the volume will change to one-third of its original volume.

Answer 2

Since you didn't mention temperature I'll have to assume this is one of those imaginary problems where temperature is (magically) held constant. In that case, the volume of a given quantity of an ideal gas at 6 atm. is 1/3 the volume of the same quantity of gas at 2 atm. Notice that 2/6 = 1/3.

Answer 3

You need to know that
P(PRESSURE) x V (VOLUME)/T(TEMPERATURE)= K (CONSTANT)
If you change P from 2 atm to 6 atm and keep T constant, then V will decrease.
If you keep the gas in the same recipient, (V constant) then T will increase in the same proportion

Answer 4

Assuming that the temperature remains constant:
pressure x volume = some constant
2 x original volume = k
6 x final volume = k

therefore the final volume would have to be 1/3 of the original for k to remain the same

Answer 5

as the pressure increases the volume decreases and the reaction shifts to the side that has the fewest number of gaseous moles present.

Answer 6

in an unchange temprature....we have the equiliburium..P1V1=P2V2..so if the pressure becomes 3 times more the changes in the volume is1/3

Answer 7

Think about:

P1*V1 = P2*V2

Solve for V2:

V2 = (P1/P2)*V1

The answer will be in terms of a fraction, (or percentage) of the initial volume, V1, which you can conveniently assign as "1" since no volume was given and is actually irrelevant.
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