Question

In a cylinder, 2.0 moles of an ideal monatomic gas initially at `1.0 xx 10^(6)` Pa and 300 K expands until its volume doulbles. Compute the work done if the expansion is (a) isothermal, (b) adiabatic and (c ) isobaric.

Answer 1

Let `P_(1) = 1 xx 10^(6) Pa = 10^(6) N//m^(2) , T_(1) = 300 K, n = 2` moles.
Final volume = 2 (initial volume) `implies V_(2) = 2V`
a. `W_("isothermal") = 2.303 nRT log (V_(2) // V_(1))`
`implies W = 2.303 xx 2 xx 8.3 xx 300 "log" 2`
`W = 3453 J`
b. `W_("adiabatic") = (nR (T_(1) - T_(2)))/(gamma - 1)`
For adiabatic process : `T_(1) V_(1)^(gamma -1) = T_(2) V_(2)^(gamma -1)`
`T_(2) = T_(1) = ((V_(2))/(V_(1)))^(gamma -1) = 300 xx ((1)/(2))^((5//3) -1) implies T_(2) = 189 K`
`W = (nRT (T_(1) - T_(2)))/(gamma - 1) = (2 xx 8.3 xx (300 - 189))/((5//3) - 1) implies W = 2764 J`
c. `W_("isobaric") = P (V_(2) - V_(1)) implies P_(1) V_(1) = nRT_(1)`
`V_(1) = (nRT_(1))/(P_(1)) = (2 xx 8.3 xx 300)/(10^(6)) implies V_(1) = 4.98 xx 10^(-3) m^(3)`
`W = P (V_(2) - V_(1)) = P (2V_(1) - V_(1)) = PV_(1)1`
`implies W = 10^(6) xx 4.98 xx 10^(-3) J implies W = 4980 J`