The temperature of 1000 gram's N2 is increased from 20°C to 100°C (at constant P). (specific heat at constant volume Cv = 5 cal/mol 0C)Then the correct statement(s) is / are

1) Q = nCp ∆T ⇒(Cv+R)∆T⇒1000285+2×120    Q = 30×103 cal ⇒30 kcal2) ∆U=nCv∆T⇒100028×5×120⇒21 Kcal3) Q=nCv∆T+P∆V for constant V, ∆V=0    Q=nCv∆T⇒21 kcal4) External work done is W = Q - ∆U=9 Kcal.