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

Step-by-step 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,

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

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

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

`(1.50 atm * 346 mL)/(318 K) = (1 atm * 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.

Answer 2

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

Step-by-step 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:

`(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=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:

`(1.50atm* 346mL)/(318K)=(1atm* V_2)/(273K)\\\\V_2=446mL`

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