A quantity of #N_2# gas originally held at 4.75 am in a 1.00L container at 26°C is transferred to a 10.0 L container at 20°C. What is the total pressure in the new container?

Also, a quantity of #O_2# gas originally at 5.25 atm and 26°C in a 5.00 L container is transferred to this same container.

1 Answer
Dec 12, 2015

Answer:

#"0.47 atm"#

Explanation:

Notice that the volume and the temperature of the gas change when going from the first container to the second container, but that the number of moles of gas remains constant.

This tells you that you can use the combined gas law equation to find the change in pressure

#color(blue)((P_1V_1)/T_1 = (P_2V_2)/T_2)" "#, where

#P_1#, #T_1#, #V_1# - the pressure, temperature, and volume of the gas in the first container
#P_2#, #T_2#, #V_2# - the pressure, temperature, and volume of the gas in the second container

Notice that the volume of the container increases by a factor of #10#, and that the temperature of the gas decreases. Both these changes allow you to precit that the pressure of the gas will be smaller in the second container.

Plug your values and solve for #P_2# - do not forget to convert the temperature from degrees Celsius to Kelvin!

#P_2 = V_1/V_2 * T_2/T_1 * P_1#

#P_2 = (1.00 color(red)(cancel(color(black)("L"))))/(10.0color(red)(cancel(color(black)("L")))) * ( (273.15 + 20)color(red)(cancel(color(black)("K"))))/( (273.15 + 26)color(red)(cancel(color(black)("K")))) * "4.75 atm"#

#P_2 = "0.4655 atm"#

You should round this off to one sig fig, the number of sig figs you have for the temperature of the gas in the second container, but I'll leave it rounded to two sig figs

#P_2 = color(green)("0.47 atm")#

I'll leave the second part of the problem, the one with the sample of oxygen, to you as practice.