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Heat and Thermodynamics question

2007 · Shift 0 · Q91

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Heat and Thermodynamics
2007 · Shift 0 · Q91

Heat and Thermodynamics question

2007 · Shift 0 · Q91

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1

When a system is taken from state

i

to state

f

along the path iaf, it is found that

Q=50

cal and

W=20cal

. Along the path

ibfQ=36cal.W

along the path

ibf

is

AIEEE 2007 Physics - Heat and Thermodynamics Question 402 English

A
$14cal$
B
$6cal$
C
$16cal$
D
$66cal$

View written solutionFree

Correct answer: B

Use the first law of thermodynamics

For any process,
$\Delta U = Q - W$
where:
- $Q$ = heat supplied to the system
- $W$ = work done by the system
From path $i \to a \to f$

Given: $Q = 50\,\text{cal}, \quad W = 20\,\text{cal}$

So,
$\Delta U_{if} = 50 - 20 = 30\,\text{cal}$
Internal energy change is path independent

Since the system goes from the same initial state $i$ to the same final state $f$ , the change in internal energy must be the same for path $i \to b \to f$ also.

Thus,
$\Delta U_{if} = 30\,\text{cal}$
For path $i \to b \to f$

Given: $Q = 36\,\text{cal}$

Using
$\Delta U = Q - W$
we get
$30 = 36 - W$
Therefore,
$W = 36 - 30 = 6\,\text{cal}$
Match with options

$W = 6\,\text{cal}$

So the correct option is B.

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