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

Answer:

1.49

Explanation:

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

Answer 2

The new volume of $H_2$ gas is required.

The new volume of the required gas is 1.49 L.

Ideal gas laws

$P_1$ = Initial pressure = 1 atm

$P_2$ = Final pressure = 2.50 atm

$T_1$ = Initial temperature = 273.15 K

$T_2$ = Final temperature = $10.00^{\circ}\text{C}=10+273.15=283.15\ \text{K}$

$V_1$ = Initial volume = 3.60 L

$V_2$ = Final volume

From the ideal gas laws we have

$\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\\\Rightarrow V_2=\dfrac{P_1V_1T_2}{T_1P_2}\\\Rightarrow V_2=\dfrac{1\times 3.6\times 283.15}{273.15\times 2.5}\\\Rightarrow V_2=1.49\ \text{L}$

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