Twelve grams of nitrogen (N2) in a steel tank are hated from 25.0 to 125ºC. (a) Howmany moles of nitrogen are present? (b) How much hat is transferred to the nitrogen?

To find out the number of moles n of nitrogen, substitute 12 g for given mass m and 28 g/mol for M (molecular weight of nitrogen) in the equation n = m/M,
n = m/M
= (12 g)/(28 g/mol)
= 0.429 mol
Thus 0.429 moles of nitrogen are present.

Since nitrogen (N₂) is a diatomic gas, therefore C_v = 5/2 R.
To obtain the heat Q which is transferred to the nitrogen, substitute 0.429 mol for n, 5/2 R for C_v and 125̊ C for T_f and 25̊ C for T_i in the equation Q = nC_v(T_f –T_i),
Q = nC_v(T_f –T_i)
= (0.429 mol) (5/2 R) (125̊ C-25̊ C)
= (0.429 mol) (5/2 ×8.31 J/mol. K) ((125 + 273) K-(25 + 273) K) (Since, R = 8.31 J/mol. K)
= (0.429 mol) (5/2 ×8.31 J/mol. K) (398 K-298 K)
= (0.429 mol) (5/2 ×8.31 J/mol. K) (100 K)
= 891 J
From the above observation we conclude that, the heat Q which is transferred to the nitrogen would be 891 J.