Question

Four identical samples of an ideal gas are taken along the paths A,B,C and D from state 1 to state 2 as shown in the PV indicator diagram. Let Q,W and ΔU represent the heat supplied, work done and change in internal energy of the gas respectively. Choose the correct option among the following :

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Solution

The correct option is **A** QA−QD=WA−WD

We know that area under P−V curve gives the work done during the process.

Work done by the gas in all four processes is positive and is in the order

WA>WB>WC>WD .....(1)

Hence, option (c) is wrong.

The change in internal energy ΔU is the same for all processes as initial and final states for all processes are the same.

Using first law of thermodynamics for each process A, B, C and D, we get

QA=ΔU+WA ......(2)

QB=ΔU+WB ......(3)

QC=ΔU+WC ......(4)

QD=ΔU+WD ......(5)

Hence, from (1) we can say that, QA>QB>QC>QD

So, option (d) is also wrong.

From equation (3) and (4),

QB−QC=WB−WC

⇒QB−WB=QC−WC

i.e option (b) is wrong.

From equation (2) and (5),

QA−QD=WA−WD

Thus, option (a) is the correct answer.

We know that area under P−V curve gives the work done during the process.

Work done by the gas in all four processes is positive and is in the order

WA>WB>WC>WD .....(1)

Hence, option (c) is wrong.

The change in internal energy ΔU is the same for all processes as initial and final states for all processes are the same.

Using first law of thermodynamics for each process A, B, C and D, we get

QA=ΔU+WA ......(2)

QB=ΔU+WB ......(3)

QC=ΔU+WC ......(4)

QD=ΔU+WD ......(5)

Hence, from (1) we can say that, QA>QB>QC>QD

So, option (d) is also wrong.

From equation (3) and (4),

QB−QC=WB−WC

⇒QB−WB=QC−WC

i.e option (b) is wrong.

From equation (2) and (5),

QA−QD=WA−WD

Thus, option (a) is the correct answer.

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