Question

A chemist conducts an experiment in which 2.0 L of hydrogen gas is collected over water at 1 atm and 298 K. Remember the pressure due to the hydrogen gas is 0.97 atm. Using the combined gas law, calculate the volume in L of the dry hydrogen gas at 273 K and a pressure of 1 atm.

Answer 1

Answer:

$V_2=1.80L$

Explanation:

Hello!

In this case, since in this problem we can see that the volume, pressure and temperature are changing, considering that the vapor pressure of water at 0 °C or also 273 K is 0.006 atm, the pressure of dry hydrogen would be 0.99 atm, thus, we can write:

$\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}$

Now, as we need the final volume V2, by solving for it, we write:

$V_2=\frac{P_1V_1T_2}{T_1P_2}$

In such a way, we plug in the given data to obtain:

$V_2=\frac{0.97atm*2.0L*273K}{298K*0.99atm}\\\\V_2=1.80L$

Which means that the underwent a compression.

Best regards!