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

Step-by-step 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,

`(P_(1)V_(1))/(T_(1)) = (P_(2)V_(2))/(T_(2))`

`(0.954 atm * 0.065 L)/(295 K) = (1 atm * 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.

Answer 2

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

Step-by-step 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:

`(P_1V_1)/(T_1)=(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:

`(725torr* 65.0mL)/(295K)=(760torr* V_2)/(273K)\\\\V_2=57.4mL`

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