Question

At conditions of 1.5 atm of pressure and 15.0 °C temperature, a gas occupies a volume of 45.5 mL. What will be the volume of the same gas at 1.1 atm and 30.0 °C?

At conditions of 1.5 atm of pressure and 15.0 °C temperature, a gas occupies a volume of 45.5 mL. What will be the volume of the same gas at 1.1 atm and 30.0 °C?

Answer 1

`"Well...what does the old combined gas equation say...?"`

Explanation:

...that `(P_1V_1)/T_1=(P_2V_2)/T_2`...for a GIVEN molar quantity of gas....the which scenario pertains here. As usual we use `"absolute temperatures"`.

And so we solve for `V_2=(P_1V_1)/T_1xxT_2/P_2`

`(1.5*atmxx45.5*mL)/(288.15*K)xx(303.15*K)/(1.1*atm)=65.3*mL`..

That volume has increased is consistent with the pressure drop, and temperature increase...the question should have specified that the gas was confined to a piston....with variable volume...

Answer 2

`"65.3 mL"`

Explanation:

You'll be using the Combined Gas Law since your question involves all three variables:

`(P_1V_1)/T_1 = (P_2V_2)/T_2`

If (`P =` pressure , `V =` volume, `T =` temperature), then we know in the equation that:

• `P_1 = "1.5 atm"`
• `P_2 = "1.1 atm"`
• `V_1 = "45.5 mL"`
• `V_2 = ?`
• `T_1 = 15^@"C"`
• `T_2 = 30^@"C"`

Before plugging in all values to solve for the volume, temperature MUST ALWAYS be in Kelvins, at least for gases. To convert Celsius to Kelvin, all you do is add `273`. So,

`T_1 = "288 K "` and `" "T_2 = "303 K"`

Now we're ready to plug and chug!

Work for calculation:

`(1.5*45.5)/288 = (1.1V_2)/303`

`0.24 = (1.1 V_2)/303`

`71.8 = 1.1 V_2`

`65.3 = V_2`

or

`V_2 = 65.3`

So that means the volume of the gas would be about `"65.3 mL"`.