Question

In a constant-volume process, 209 j of energy is transferred by heat to 1.00 mol of an ideal monatomic gas initially at 300 k. find (a) the work done on the gas, (b) the increase in internal energy of the gas, and (c) its final temperature.

Answer 1

 (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