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
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
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
#1.004nRT = nR(T + 1)#
Then, divide by
#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.