0.35 g of Nonyne gas was exposed to a temperature of 29 degrees C and a volume of 5.7 gallons. How much pressure does this exert?

Question

Luna Chen · Accepted Answer

Are you sure #"nonyne"# is a gas.........?

I don't have an Aldrich or Acros catalogue handy. But I am pretty sure that the isomeric nonynes would be distillable liquids (if they are available commercially!). I will assume that you speak of acetylene, #HC-=CH#, which is certainly a room temperature gas, and answer the question on this basis. There is also another assumption I must make; that is that you speak of #"US gallons"#. So far as I know, and I try to avoid the use of whack units, #"1 US gallon"# #=# #3.79*L#. On the other hand, you might speak of #"Imperial Gallons"# #=# #4.54*L# (I would certainly use the latter unit here in the UK!). This confusion nicely illustrates why scientists (and syllabuses) should use non-ambiguous, standard units. #"Moles of acetylene"=(0.35*g)/(26.04*g*mol^-1)=1.34xx10^-2*mol#. And using the #"Ideal Gas Equation: "P=(nRT)/V# #=(1.34xx10^-2*molxx0.0821*(L*atm)/(K*mol)xx302*K)/(5.7*"gallons"xx3.79*L*"gallon"^-1)# #=1.54xx10^-2*atm-=11.7*mm*Hg#. In any case, the entire inquiry is highly dubious.

Serenity Allen · Answer

undefined Use the ideal gas law to calculate pressure. $PV = nRT$. Given dataUse the ideal gas law to calculate pressure. $PV = nRT$. Given  $n = \frac{0.35 \, \text{g}}{114 \, \text{g/mol}}$, $R = 0.0821 \, \text{L}\cdot\text{atm/(mol}\cdot\text{K)}$, $T = 29 \, \text{°C} + 273.15$, $V = 5.7 \, \text{gallons}$. Calculate pressure in atm.