(a) the work done on the gas
From the definition of work in thermodynamics,
W = integral (PdV)
It is said that the process above is done at constant volume. Therefore, the value of dV is zero which would result to
W = integral (PdV) = 0
(b) the increase in internal energy of the gas
From the first law of thermodynamics,
ΔU = Q + W
where U is the change in internal energy, Q is the heat and W is the work which is zero in this case
ΔU = Q = 209 J
(c) its final temperature
at constant volume,
Q = nCv(T2 - T1)
209 = 1 (3R/2) (T2 - 300)
T = 316.76 K
