For a certain gas in a closed container, the pressure has been raised by #0.4%#, and the temperature was raised by #"1 K"#. What temperature did the gas start at?

#a)# #"250 K"#
#b)# #"200 K"#
#c)# #"298 K"#
#d)# #"300 K"#

1 Answer
Jul 29, 2017

This is an impossible question. However, if we assume that the question should have written that the vessel is rigid, then I get an initial temperature of #"250 K"#.


Well, as usual, when you see pressure, temperature, and "closed vessel" in the same sentence, we assume ideality...

#PV = nRT#

  • #P# is pressure in #"atm"#.
  • #V# is volume in #"L"#.
  • #n# is mols of ideal gas.
  • #R = "0.082057 L"cdot"atm/mol"cdot"K"# is the universal gas constant.
  • #T# is temperature in #"K"# (as it must be! Why?).

The closed vessel means very little to us; all it says is that the mols of gas are constant. It says nothing about the volume of the vessel being constant, as it could very well be a big fat balloon.

We apparently are given:

#P -> 1.004P#

#T -> T + 1#

#n -> n#

#V -> ??? xx V#

Substitute to get:

#1.004P cdot V = nR(T+1) = nRT + nR#,

where the written variables are all for the initial state and we interpret the question to mean a closed AND rigid vessel. So, we assume that #??? = 1#.

Note that since #PV = nRT#, we can now write:

#1.004nRT = nR(T + 1)#

Then, divide by #nR# to get:

#1.004T = T + 1#

#=> 0.004T = 1#

#=> color(blue)(T = 1/0.004 = "250 K")#

which is one of the given answer choices. It doesn't mean the question can't be revised, but that is probably what the question actually meant.