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
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")