If the volume of the container is #"1.00 L"#, what is the change in mols of gas required to change the pressure from #"1.00 atm"# to #"0.920 atm"# if the temperature was also dropped from #25^@ "C"# to #22^@ "C"#? (#R = "0.082057 L"cdot"atm/mol"cdot"K"#)

Question

Olivia Anton · Accepted Answer

I get about #Deltan = n_2 - n_1 = "0.00289 mols"#. Since you are given: we know that the volume had to have remained constant (we only got one value for it). Also, since we were given pressure, temperature, volume, and #R#, it's a good hint that we are using the ideal gas law: #PV = nRT# Since we have the initial and final values for pressure and temperature, we have two variations on the ideal gas law: #P_1V = n_1RT_1#
#P_2V = n_2RT_2# So finding "the mols" should mean finding the change in the mols, #n_2 - n_1 = Deltan#. A reasonable way to do this is to solve for the initial #"mol"#s: #n_1 = (P_1V)/(RT_1)# #= (("1.00 atm")("1.00 L"))/(("0.082057 L"cdot"atm/mol"cdot"K")("25 + 273.15 K"))# #=# #"0.0409 mols"# So, #n_2# is gotten as follows: #n_2 = (P_2V)/(RT_2)# #= (("0.920 atm")("1.00 L"))/(("0.082057 L"cdot"atm/mol"cdot"K")("22 + 273.15 K"))# #=# #"0.0380 mols"# So, if you wanted the change in #"mol"#s, you would get: #color(blue)(Deltan)# #=# #"0.0409 - 0.0380 mols"# #=# #color(blue)("0.00289 mols")#

Wyatt Atkins · Answer

undefined Using the ideal gas law, we can rearrange it to solve for the change in moles of gas:

Δn = Δ(PV) / (R * ΔT)

Where Δn is the change in moles of gas, Δ(PV) is the change in pressure times volume (final pressure times final volume minus initial pressure times initial volume), R is the gas constant, and ΔT is the change in temperature in Kelvin.

Δ(PV) = (0.920 atm * 1.00 L) - (1.00 atm * 1.00 L) = -0.08 atm * L

ΔT = (25 + 273) K - (22 + 273) K = 3 K

Now, substitute these values into the formula:

Δn = (-0.08 atm * L) / (0.082057 L·atm/mol·K * 3 K) ≈ -0.032 mol

So, the change in moles of gas required to change the pressure from 1.00 atm to 0.920 atm when the temperature is dropped from 25°C to 22°C is approximately -0.032 moles.