An insulated container has 5 moles of an ideal diatomic gas at temperature T. When Q heat is added to the gas, 3 moles of the gas are dissociated into atoms, (the temperature remaining same) then [R is gas constant] a. Q = RT b. Q = 1.5 RT c. Q = 2 RT d. Q = 2.5 RT
so when we add Q amount of heat 3 moles of diaatomic breaks to monoatomic...so we have 6 moles monoatomic gas....so total dU= finalU- initial U....... = (6*3/2 RT + 2*5/2 RT ) - 5*5/2 RT = 14RT - 12.5 RT = 1.5 RT....
so the answer is b, 1.5 RT.... got it...it's a simple problem..
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