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7 MOLES OF TETRA ATOMIC NON LINEAR GAS A AT 10 ATM AND T K ARE MIXED WITH 6 MOLES OF ANOTHER GAS B AT T/3 K AND 5 ATM IN A CLOSED RIGID VESSEL WITHOUT ENERGY TRANSFER WITH SORROUNDINGS IF THE FINAL TEMPRATURE OF THE MIXTURE WAS 5T/6 K , THEN GAS B IS? DIATOMIC OR TRIATOMIC ?
Cv of gas A = 3/2 R + 3/2 R + 6R = 9RLet Cv of gas B = CvHeat lost by gas A = 7 x 9R (T- 5T/6)Heat gained by gas B = 6 Cv (5T/6 – T/3)63 R(T/6) = 6 Cv (T/2)Cv=7 R/2therefore, gas is diatomic = 3R/2 + R + R = 7 R/2
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