2 moles of helium (He) are pumped from a gas canister into a weather balloon. The balloon at this point has a volume V1.

1 mole of nitrogen gas (N2) is added at the same temperature and pressure. The volume of the balloon changes to V2.

How will the new volume of the balloon compare with the old volume (V1)?
1) 2 × V1
2) 0.50 × V1
3) None of these
4) There will be no change in volume.
5) 1.5 × V1
6) 0.67 × V1

Won't it be V1*(3/2)?

P1V1 = n1RT but P, R and T are constants so V1 = kN1 and V2 = kN2
(V1/V2) = (n1/n2)

To determine the new volume of the balloon after adding 1 mole of nitrogen gas, we can use the ideal gas law, which states:

PV = nRT

Where:
P is the pressure of the gas
V is the volume of the gas
n is the number of moles of gas
R is the ideal gas constant
T is the temperature of the gas

We know that the temperature and pressure remain constant, as mentioned in the question. Therefore, we can rearrange the ideal gas law equation as follows:

V = (nRT) / P

Since we are comparing the volume before and after adding nitrogen gas, we can use the equation:

(V2) = (n1RT) / P

where n1 = 2 moles of helium

Now, let's solve for V2:

V2 = (2 moles * R * T) / P

Next, we need to calculate the volume of the balloon before adding nitrogen gas, denoted as V1. Given that the initial volume is V1 and there are 2 moles of helium:

V1 = (2 moles * R * T) / P

To determine the new volume of the balloon after adding 1 mole of nitrogen gas, we need to calculate V2/V1:

(V2/V1) = [(2 moles * R * T) / P] / [(2 moles * R * T) / P]

As we can see, all the terms cancel out, resulting in:

(V2/V1) = 1

This means that the new volume of the balloon (V2) is the same as the old volume (V1). Therefore, the correct answer is:

4) There will be no change in volume.