Question

15.0 L of an ideal gas at 298 K and 3.36 atm are heated to 383 K with a new pressure of 5.60 atm. What is the new volume in liters

Answer 1

Answer:

$V_2=11.6L$

Explanation:

Hello!

In this case, since we volume, pressure and temperature which are all changing, we can use the combined ideal gas law to write:

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

Thus, since the final volume V2 is required, 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:

$V_2=\frac{3.36atm*15.0L*383K}{298K*5.60atm}\\\\V_2=11.6L$

Which means that the process compressed the gas.

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