If 18.75 mol of helium gas is at 10.0 degree C and a gauge pressure of 0.350 atm, a. Calculate the volume of the helium gas under these conditions. b. Calculate the temperature if the gas is compressed to precisely half the volume at a gauge pressure of 1.00 atm.

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2poclâ‚(g) ⇜ 2 pclâ‚(g) + oâ‚‚(g) â†gâ° = (2 mol)î”gfâ°(pclâ‚) – (2 mol)î”gfâ°(poclâ‚) = (2 mol)(–270 –(–502) kj/mol) â†gâ° = + 464 kj = â†hⰠ– tî”sâ° î”sâ° = 179 j/k ⇒ â†hâ° = â†gâ° + (298 k)(179 j/k) = –411 kj assuming that î”hâ° and î”sâ° do not depend on temperature, the temperature at which â†gâ° = 0 is where t = â†hâ°/î”sâ° = (–411 kj)/(0.179 j/k) = 2294 k. at that temp. keq = 1 for the reaction.