SATHEE: Heat and Thermodynamics 5 Question 41

Heat and Thermodynamics 5 Question 41

42. A thermodynamic system is taken from an initial state $i$ with internal energy $U_{i} = 100$ $J$ to the final state $f$ along two different paths iaf and $i b f$ , as schematically shown in the figure. The work done by the system along the paths $a f, i b$ and $b f$ are $W_{a f} = 200 J, W_{i b} = 50$ $J$ and $W_{b f} = 100 J$ respectively. The heat supplied to the system along the path $i a f, i b$ and $b f$ are $Q_{i a f}, Q_{i b}$ and $Q_{b f}$ respectively. If the internal energy of the system in the state $b$ is $U_{b} = 200 J$ and $Q_{i a f} = 500 J$ , the ratio $Q_{b f} / Q_{i b} i$

(2014 Adv.)

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Answer:

Correct Answer: 42. (2)

Solution:

$W_{i b f} = W_{i b} + W_{b f} = 50 J + 100 J = 150 J$

$\begin{aligned} W_{i a f} & = W_{i a} + W_{a f} = 0 + 200 J = 200 J \\ Q_{i a f} & = 500 J \\ Δ U_{i a f} & = Q_{i a f} - W_{i a f} \\ = 500 J - 200 J \\ = 300 J = U_{f} - U_{i} \\ So U_{f} & = U_{i a f} + U_{i} \\ = 300 J + 100 J = 400 J \\ Δ U_{i b} & = U_{b} - U_{i} \\ = 200 J - 100 J = 100 J \\ Q_{i b} & = Δ U_{i b} + W_{i b} \\ = 100 J + 50 J = 150 J \\ Q_{i b f} & = Δ U_{i b f} + W_{i b f} = Δ U_{i a f} + W_{i b f} \\ = 300 J + 150 J = 450 J \end{aligned}$

So, the required ratio

$\begin{aligned} \frac{Q_{b f}}{Q_{i b}} & = \frac{Q_{i b f} - Q_{i b}}{Q_{i b}} \\ = \frac{450 - 150}{150} = 2 \end{aligned}$

Heat and Thermodynamics 5 Question 41

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Heat and Thermodynamics 5 Question 41

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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