Question

1. A volume of 120.0 liters of a gas is prepared at a pressure of 400.0 mm Hg and a temperature

Answer 1

Answer:

$V_2=2487.9L$

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to calculate the final volume by using the combined ideal gas:

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

In such a way, by solving for the final volume, V2, we obtain:

$V_2=\frac{P_1V_1T_2}{P_2T_1}$

Now, by plugging in the pressures, temperatures in Kelvins and initial volume, we will obtain:

$V_2=\frac{(400.0mmHg)(120.0L)(283.15K)}{(20.0mmHg)(273.15K)}\\\\V_2=2487.9L$

Regards!