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?

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?

Jeremiah Adams · Accepted Answer

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

...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#..

The temperature rise and pressure drop are consistent with the volume increase. The question ought to have made it clear that the gas was contained within a piston and had a variable volume.

Avery Adams · Answer

#"65.3 mL"#

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:

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"#.

Cooper Anderson · Answer

undefined To solve this problem, you can use the combined gas law, which states that the ratio of the initial pressure, volume, and temperature of a gas to its final pressure, volume, and temperature is constant when the amount of gas is held constant. The formula is: (P1 * V1) / (T1) = (P2 * V2) / (T2). Rearranging the formula to solve for V2 gives: V2 = (P1 * V1 * T2) / (P2 * T1). Plugging in the given values, you can calculate the volume of the gas at the new conditions: V2 = (1.5 atm * 45.5 mL * 303.15 K) / (1.1 atm * 288.15 K) = 69.2 mL. Therefore, the volume of the gas at 1.1 atm and 30.0 °C will be approximately 69.2 mL.