500.0 L of gas are prepared at 0.921 atm pressure and 200.0°C. The gas is placed into a tank under high pressure. When the tank cools to 20.0°C, the pressure is 30.0 atm. What is the volume of the gas under these conditions?

1 Answer
Jacob T.
Jul 4, 2018

#"V" = 24.8 color(white)(l) "L"#

Explanation:

The ideal gas law suggests the relationship

#P * V = n * R * T#

that relates

  • #P#, the pressure of the gas
  • #V#, the volume the gas occupies
  • #n#, the number of moles of gas in the closed container
  • #T#, the temperature of the gas in #color(purple)("degrees Kelvin")#, and
  • #R# the ideal gas constant.

Let #n_1#, #P_1#, #V_1#, and #T_1# resembles the respective properties of the gas before the change and #n_2#, #P_2#, #V_2#, and #T_2# properties after the change.

  • #n_1 = (R * T_1) / (P_1 * V_1)#

  • #n_2 = (R * T_2) / (P_2 * V_2)#

The amount of gas present in the container, #n#, stays the same during this process. #n_1 = n_2# and therefore

#(R * T_1) / (P_1 * V_1) = n = (R * T_2) / (P_2 * V_2)#

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

Multiplying both sides of the equation by #(P_1 * V_2) / (T_1)# gives the relationship between the volume of the gas before and after the change:

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

Therefore

#V_2 = V_1 * ((T_2) / (T_1) * (P_1) / (P_2))#
#color(white)(V_2) = 500.0 color(white)(l) "L" xx ((200 + 273.15) color(white)(l) color(red)(cancel(color(purple)(K)))) / ((20 + 273.15) color(white)(l) color(red)(cancel(color(purple)(K)))) * (0.921 color(white)(l) color(red)(cancel(color(black)("atm")))) / (30.0 color(white)(l) color(red)(cancel(color(black)("atm"))))#
#color(white)(V_2) = 24.8 color(white)("L")#

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