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Question

In the given graph an ideal gas change its state from A to state C by two paths ABC and AC.

The internal energy of gas at state B is 20 J. The amount of heat (in J) supplied to the gas to go from A to B is 2×X.Then value of x is;
217800_015f3e60104d4b02b24d1bdfc050fb0a.JPG

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Solution

QBA=UBA+WBA
It is an isochoric process so work done will be zero.
Now,
QBA=UAUB
TA=20nR------(PV=nRT)
UA=1.5nRTA=30 Joule
SO QBA=3020=10 Joule

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