Another gas problem?

Question

A 2.5 g sample of He is held in a 3.000 L closed container at 20.0 degrees celcius. The temperature is increased to 60.0 degrees celcius. If the final volume is 3.000 L and the final pressure is 1.874 atm, what was the initial pressure of the gas?

Does the mass play any role in this problem?

Answer 1

Ok, let me explain the concept behind this better.

You're going to use the Ideal Gas Law, PV=nRT

R will always be constant, so if you're comparing a before and after situation, you can drop it.

And being that no matter which change you make to a gas, it will have to obew the Ideal Gas Law, you can make the following relationship:

P1V1/n1T1 = P2V2/n2T2

If you don't change the number of moles, then n1 and n2 just disappear from the equation.

Same thing happens if the temperature remains the same, or the volume or pressure.

You just need to adjust this relationship for the situation you're examining.

Answer 2

Use the balanced equation to calculate the range of moles of H2 gas that are produced by the reaction with that categorical quantity of Al. you keep in mind that at STP a million mole of any gas occupies a quantity of twenty-2.4 litres, so use the rigidity and temperature files so that you could calculate the quantity of that a lot H2 gas by using the courting. P1V1/T1 = P2V2/T2 (that you'll derive from the acceptable gas regulation PV = nRT in case you quite opt for to).

Answer 3

I forgot the formula, but I think since the volume is constant and the pressure and temperature change,
P1V1/T1 = P2V2/T2
P1/293K = 1.874/ 333K
P1 = 1.874 x 293 / 334
= 1.64Atm

Answer 4

PV = nRT for both the intial state and the final state.

Since V has not changed:

Pf/Pi = Tf/Ti, make sure that you use absolute temperatures.

Answer 5

Another gas problem! Stop eating so many beans, *****!