Question

A sample of argon (Ar) gas occupies 65.0 mL at 22.0°C and 725 torr. What is the volume of this Ar gas at STP? Enter your answer in the provided box.

Answer 1

Answer: The volume of argon gas at STP is 57.4 mL

Explanation:

STP conditions are:

Pressure of the gas = 1 atm = 760 torr

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:

$P_1=725torr\\V_1=65.0mL\\T_1=22^oC=(22+273)K=295K\\P_2=760torr\\V_2=?\\T_2=273K$

Putting values in above equation, we get:

$\frac{725torr\times 65.0mL}{295K}=\frac{760torr\times V_2}{273K}\\\\V_2=57.4mL$

Hence, the volume of argon gas at STP is 57.4 mL

Answer 2

Explanation:

The given data is as follows.

     $V_{1}$ = 65.0 mL = 0.065 L (as 1 ml = 0.001 L),      

     $T_{1}$ = $22.0^{o}C$ = (22 + 273) K = 295 K,

      $P_{1}$ = 725 torr = 0.954 atm           (as 1 torr = 0.00131579 atm),

     $V_{2}$ = ?,   $T_{2}$ = 273 K,

      $P_{2}$ = 1 atm

And, according to ideal gas equation,

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

           $\frac{0.954 atm \times 0.065 L}{295 K} = \frac{1 atm \times V_{2}}{273 K}$

            $V_{2}$ = 0.0574 L

As, 1 L = 1000 ml. So, 0.0574 L = 57.4 ml.

Thus, we can conclude that the volume of this Ar gas at STP is 57.4 L.