Question

A 6.25 L sample of gas is collected at 298 K and 2.50 atm. What is the volume of the sample at STP?

Answer 1

The volume of a 6.25 L gas sample at 298 K and 2.50 atm at STP is calculated using the combined gas law, which results in a final volume of approximately 14.49 L.

The question involves using the combined gas law to determine the volume of a gas sample at STP (standard temperature and pressure) given its initial conditions. STP is defined as 0°C (273.15 K) and 1 atm pressure. We use the combined gas law which is expressed as (
`P_1`*
`V_1`) /
`T_1` = (
`P_2`*
`V_2`) /
`T_2` , where P is the pressure, V is the volume, and T is the temperature in Kelvin.

For the given sample with an initial volume (
`V_1`) of 6.25 L, initial pressure (
`P_1`) of 2.50 atm, and initial temperature (
`T_1`) of 298 K, we want to find the final volume (
`V_2`) at standard temperature (
`T_2` = 273.15 K) and standard pressure (
`P_2`= 1 atm).

We rearrange the formula to solve for
`V_2`:

`V_2`= (
`P_1` *
`V_1`*
`T_2`) / (
`P_2`*
`T_1`)

`V_2`= (2.50 atm * 6.25 L * 273.15 K) / (1 atm * 298 K)

`V_2`= (4319.6875 L*K) / (298 K) ≈ 14.49 L

So, at STP, the volume of the sample will be approximately 14.49 L.