Can someone please solve this problem

Question

A hot air balloon is filled with 1.19×106 L of an ideal gas on a cool morning ( 11 ∘C). The air is heated to 101 ∘C. What is the volume of the air in the balloon after it is heated? Assume that none of the gas escapes from the balloon. volume:

Answer 1

The ideal gas law says: PV=nRT were P=pressure, V=volume, n=amount of substance, R=ideal gas constant, and T=temperature.

We know the volume of the gas (1.19*10^6 L) at the temperature of 11 C (which is 284 K). The pressure in the balloon is equal to the barometric pressure of the outside air (because otherwise the balloon would stretch or shrink), and the amount of gas and the ideal gas constant do not change in this situation. We can therefore substitute values in the ideal gas law as follows: P*1.19*10^6 = nR* 284.

We can now take the three constants (P, n, and R) and calculate a combined value for them, as follows:

P/(nR) = 284/1.19*10^6 = 2.39*10^-4 = T/V

Now, using the new value of T=374, we can substitute 2.39*10^-4 = 374/V.

Multiplying both sides by V and dividing both sides by 2.39*10^-4, we can say V = 374/(2.39*10^-4) = 1.567*10^6.

Alternatively, we could have said, V = T / 2.39*10^-4 = 374/2.39*10^-4 = 1.567*10^6.

Expressing the solution to the same accuracy as the data in the problem, we would say that the final volume of the balloon is 1.58 * 10^6 L.