Question

A sample of 02 gas occupies 346 mL at 45°C and 1.50 atm. What is the volume of this O2 gas sample at STP? Enter your answer in the provided box. L L

Answer 1

Answer: The volume of oxygen gas at STP is 446 mL

Explanation:

STP conditions are:

Pressure of the gas = 1 atm

Temperature of the gas = 273 K

To calculate the volume when temperature and pressure has changed, we use the equation given by combined gas law. The equation follows:

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

where,

$P_1,V_1\text{ and }T_1$ are the initial pressure, volume and temperature of the gas

$P_2,V_2\text{ and }T_2$ are the final pressure, volume and temperature of the gas

We are given:

$P_1=1.50atm\\V_1=346mL\\T_1=45^oC=(45+273)K=318K\\P_2=1atm\\V_2=?\\T_2=273K$

Putting values in above equation, we get:

$\frac{1.50atm\times 346mL}{318K}=\frac{1atm\times V_2}{273K}\\\\V_2=446mL$

Hence, the volume of oxygen gas at STP is 446 mL

Answer 2

Explanation:

The given data is as follows.

     $V_{1}$ = 346 mL,     $T_{1}$ = $45.0^{o}C$ = (45 + 273) K = 318 K,

      $P_{1}$ = 1.50 atm,   $V_{2}$ = ?,   $T_{2}$ = 273 K,

      $P_{2}$ = 1 atm

And, according to ideal gas equation,

               $\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}$

Now, putting the given values into the above formula and we will calculate the final volume as follows.

          $\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}$

           $\frac{1.50 atm \times 346 mL}{318 K} = \frac{1 atm \times V_{2}}{273 K}$

            $V_{2}$ = 445.56 mL

Thus, we can conclude that the volume of this $O_{2}$ gas sample at STP is 445.56 mL.