A 2.00 mol sample of helium gas initially at 300 K and 0.400 atm is compressed isothermally to 1.20 atm. Noting that the helium behaves as an ideal gas, find (a) the final volume of the gas, (b) the work done on the gas and (c ) the energy transferred by heat.

Explanation

The final volume will be one-third of the original, for the temperature to be constant, the sample must liberate as many joules by heat as it takes in by work.
The ideal gas law can tell us the original and final volumes. The negative integal of PdV will tell us the work input, and the first law of thermodynamics will tell us the energy output by heat.
a. Rearranging `PV = nRT`, we get `V_(i) = (nRT)/(P_(i))`
The initial volume is
`V_(i) = ((2.00 mol)(8.314 J mol K)(300 K))/((0.400 atm)(1.013 xx 10^(5) Pa//atm)) (1Pa)/(N//m^(2))`
For isothermal compression, PV is constant, so `P_(i) V_(i) = P_(f) V_(f)`
`V_(f) - V_(i) ((P_(i))/(P_(f))) = (0.123 m^(3)) ((0.400 atm)/(1.20 atm)) = 0.0410 m^(3)`
b. `W = - int PdV = - int (nRT)/(V) dV = - nRT 1n ((V_(f))/(V_(i)))`
`= - (4988 J) 1n ((1)/(3)) = + 5.48 kJ`
c. The ideal gas keeps constant temperature so `Delta E_(i n t) = 0`
`= Q + W` and the heat is `Q = - 5.48 kJ`