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Question

An ideal gas is taken from state 1 to state 2 through optional path A, B , C & D as shown in P-V diagram. Let Q, W and U represent the heat supplied, work done & internal energy of the gas respectively. Then 
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A
QBWB>QCWC
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B
QAQD=WAWD
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C
WA<WB<WC<WD
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D
QA>QB>QC>QD
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Solution

The correct options are
B $$ Q_A - Q_D = W_A -W_D$$
D $$ Q_A > Q_B > Q_C > Q_D $$
Work done is equal to the area under the P-V curve.
From the graph,  area under path A is the largest whereas under path c is the smallest.
Thus $$W_A>W_B>W_C>W_D$$             ............(1)
From Ist law of thermodynamics,    $$ Q= \Delta U+  W$$     ........(2)
As $$\Delta U $$ is a state function, thus $$\Delta U $$ is same for all paths.
From (1) & (2),       $$Q_A>Q_B>Q_C>Q_D $$
Also  for path A and D,  $$Q_A=\Delta U+ W_A$$   and    $$Q_D=\Delta U+W_D$$
Subtracting these two,  $$\implies  Q_A-Q_D= W_A-W_D$$
Similarly,   $$Q_B-W_B=Q_C-W_C$$

Physics

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