You have a sample of neon gas at a certain pressure, volume, and temperature. You double the volume, double the number of moles of neon, and double the Kelvin temperature. How does the final pressure (Pf) compare to the original pressure (Po) ?

Question

Jeremiah Adams · Accepted Answer

Additionally, the final pressure will double. If you use the ideal gas law equation, #PV = nRT#, to express the initial and the final states of your gas sample, you can write something like this  #P_1V_1 = n_1 * R * T_1# #-># the initial state; (1)
#P_2V_2 = n_2 * R * T_2# #-># the final state;  You also know that the volume doubles, which means that #V_2 = 2 * V_1#, the number of moles doubles, #n_2 = 2 * n_1#, and the temperature doubles, #T_2 = 2 * T_1#. Use these equalities into the equation that describes the final state #P_2 * 2 * V_1 = 2 * n_1 * R * 2 * T_1# (2) The result of dividing equation (1) by equation (2) is #P_1/(2 * P_2) = (n_1 * R * T_1)/(2 * n_1 * R * 2 * T_1)# #P_1/( 2 * P_2) = 1/4 => 2 * P_2 = 4 * P_1 => P_2 = 2 * P_1# Lastly, the answer makes sense since both the temperature and the number of moles will double the pressure, but one of these effects will be neutralized by the volume doubling as well. As a result, the pressure also doubles. The pressure would have increased four times if the volume had remained constant: twice as much because the temperature doubled and twice as much because there were twice as many moles.

Noah Aldrich · Answer

undefined The final pressure (Pf) will be the same as the original pressure (Po). This is described by the ideal gas law, which states that pressure is directly proportional to the number of moles of gas (n) and the temperature (T), and inversely proportional to the volume (V). Therefore, if the volume, number of moles, and temperature are all doubled, the ratio of $ \frac{n \cdot T}{V} $ remains constant, resulting in no change in pressure.