Question

1.00 mole of an ideal gas occupies a volume of 22.4 L at 0 C and 760 mmHg (STP). It is cooled to -41 C while the pressure is increased to 805 mmHg. What is the new volume of the gas in liters?

Answer 1

Answer:

$V_2=18.0L$

Explanation:

Hello there!

In this case, according to the given information, we infer that we need the combined gas law in order to solve this problem:

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

Thus, by suing the temperatures in Kelvins, we can solve for the final volume as shown below:

$V_2 =\frac{P_1V_1T_2}{T_1P_2}}\\\\V_2 =\frac{(760mmHg)(22.4L)(232K)}{(273K)(805mmHg)} \\\\V_2=18.0L$

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