Question

a balloon is filled with 3.60 L of H2 gas at STP. If the balloon is taken into the sea where the pressure is 2.50 atm and the temperature is 10.00c, what is the new volume of H2 gas in liter?

Answer 1

The new volume of the hydrogen gas in the balloon is 1.5 L.

Explanation:

At STP, Temperature, T1 is 0° C = 0 + 273 K = 273 K

Pressure, P1 = 1 atm

Here given volume, V1 = 3.60 L

Balloon at sea,

Pressure, P2 = 2.5 atm

Temperature, T2 = 10° C = 10 + 273 = 283 K

Volume, V2 = ?

Here, we have to use the equation,

$\frac{P1V1}{T1} = \frac{P2V2}{T2}$

We have to rearrange the equation to get V2 as,

$V2 = \frac{P1V1T2}{T1P2}$

Now plugin the values as,

V2 = $\frac{1 atm \times 3.6 L \times 283 K}{273 K \times 2.5}$

   = 1.5 L

So the new volume of the hydrogen gas is 1.5 L.

Answer 2

Answer:

1.49

Explanation:

V2= (1 atom)(3.60L)(283K) over (2.50 atm)(273K) = 1.49 L