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Mechanical Properties Of Fluids

Question
CBSEENPH11020544

Figure below shows two paths that may be taken by a gas to go from a state A to a state C. 


In process AB, 400 J of heat is added to the system and in process BC, 100 J of heat is added to the system. The heat absorbed  by the system in the process AC will be

  • 380 J

  • 500 J

  • 460 J

  • 300 J

Solution

C.

460 J

Since, initial and final points are same

So comma
increment straight U subscript straight A rightwards arrow straight B rightwards arrow straight C end subscript space equals space increment straight U subscript straight A rightwards arrow straight C end subscript space.. space left parenthesis straight i right parenthesis
Also space straight A rightwards arrow straight B space is space isochoric space process
So space dW subscript straight A rightwards arrow straight B end subscript space equals space 0
and space dQ space equals space dU plus space dW
so comma space dQ subscript straight A rightwards arrow straight B end subscript space equals space dU subscript straight A rightwards arrow straight B end subscript space equals space 400 space straight J
Next space straight B rightwards arrow straight C space space is space isobaric space process
so comma space dQ subscript straight B minus straight C end subscript space equals space dU subscript straight B rightwards arrow straight C end subscript space plus dW subscript straight B rightwards arrow straight C end subscript
equals space dU subscript straight B rightwards arrow straight C end subscript plus straight p increment straight V subscript straight B rightwards arrow straight C end subscript
rightwards double arrow 100 space equals space dU subscript straight B rightwards arrow straight C end subscript plus 6 space straight x space 10 to the power of 4 left parenthesis 2 space straight x space 10 to the power of negative 3 end exponent right parenthesis
rightwards double arrow dU subscript straight B rightwards arrow straight C space end subscript space equals space 100 minus 120 space equals space minus 20 space straight J
From space eq space left parenthesis straight i right parenthesis space
because space increment straight U subscript straight A rightwards arrow straight B rightwards arrow straight C end subscript space equals space increment straight U subscript straight A rightwards arrow straight C end subscript
space increment straight U subscript straight A rightwards arrow straight B end subscript space plus increment straight U subscript straight B rightwards arrow straight C end subscript space equals space increment straight U subscript straight A rightwards arrow straight C end subscript minus dW subscript straight A rightwards arrow straight C end subscript
400 plus space left parenthesis negative 20 right parenthesis space equals space dQ subscript straight A rightwards arrow straight C end subscript
minus left parenthesis straight p increment straight V subscript straight A space plus area space straight o increment ABC right parenthesis
rightwards double arrow space dQ subscript straight A rightwards arrow straight C end subscript space equals space 380 plus open parentheses 2 space straight x space 10 to the power of 4 space straight x space 2 space straight x space 10 to the power of negative 3 end exponent plus 1 half space straight x space 2 straight x space 10 to the power of minus straight x space 4 space straight x space 10 to the power of 4 close parentheses
equals space 380 space plus space left parenthesis 40 plus 40 right parenthesis
dQ subscript straight A rightwards arrow straight C end subscript space equals space 460 space straight J