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?

...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...

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"#.