Question

A cylinder of volume 25.0L is filled with He gas at a pressure of 5.0atm and 450K. How many balloons of capacity of 1.0L can be filled using the same He gas at pressure of 1.0atm and temperature of 300K​

Answer 1

Answer:

When an ideal gas is compressed adiabatically \left(Q=0\right), work is done on it and its temperature increases; in an adiabatic expansion, the gas does work and its temperature drops. Adiabatic compressions actually occur in the cylinders of a car, where the compressions of the gas-air mixture take place so quickly that there is no time for the mixture to exchange heat with its environment. Nevertheless, because work is done on the mixture during the compression, its temperature does rise significantly. In fact, the temperature increases can be so large that the mixture can explode without the addition of a spark. Such explosions, since they are not timed, make a car run poorly—it usually “knocks.” Because ignition temperature rises with the octane of gasoline, one way to overcome this problem is to use a higher-octane gasoline.

Another interesting adiabatic process is the free expansion of a gas. (Figure) shows a gas confined by a membrane to one side of a two-compartment, thermally insulated container. When the membrane is punctured, gas rushes into the empty side of the container, thereby expanding freely. Because the gas expands “against a vacuum” \left(p=0\right), it does no work, and because the vessel is thermally insulated, the expansion is adiabatic. With Q=0 and W=0 in the first law, \text{Δ}{E}_{\text{int}}=0, so {E}_{\text{int}}{}_{i}={E}_{\text{int}}{}_{f} for the free expansion.

The gas in the left chamber expands freely into the right chamber when the membrane is puncture

Answer 2

Answer:

83 balloons can be filled using the same He gas in given conditions.

Explanation:

Provided that:

Initial volume (V1) = 25.0 L

Initial pressure (P1) = 5.0 atm

Initial temperature (T1) = 450 K

And;

Final pressure (P2) = 1.0 atm

Final temperature (T2) = 300 K

Final Volume = V2 L (Let's suppose)

From the Boyle's law we know: PV = Constant

From Charles law we know: V/T = Constant

From Gay-Lussac's law we know: P/T = C

Combining these three laws we get the combined gas law:

$PV = Constant \\ \frac{V}{T} = Constant \\ \frac{P}{T} = Constant \\ \\ \frac{PV}{T} =Constant \: \: ...equation(i) \\ or \: \frac{P1 \: V1}{T1} = \frac{P2 \: V2}{T2} \: \: ...equation(ii)$

Hence from equation (ii) we can calculate the final Volume V2 :

$\frac{P1 \: V1}{T1} = \frac{P2 \: V2}{T2} \\ or \: V2 = \frac{P1 \: V1 \:T2 }{P2 \: T1} \\ \\ so \: V2 = \frac{5 \times 25 \times 300 }{1 \times 450} \: litre \\ = 83.333 \: litre$

Hence we get the final volume of Helium gas is 83.333 L

Now we have to fill balloons of 1.0 L in these final conditions.

So the number of balloons (n) we can fill up with this volume of Helium gas:

$n = \frac{83.333}{1.0} \: number \: of \: ballons \\ n = 83 \: ballons \: (Taking \: Integer )$

Hence we get; 83 balloons can be filled using the same He gas in given conditions.