SATHEE: Heat and Thermodynamics 6 Question 13

Heat and Thermodynamics 6 Question 13

17. Now consider the partition to be free to move without friction so that the pressure of gases in both compartments is the same. Then total work done by the gases till the time they achieve equilibrium will be

(2014 Adv.)

(a) $250 R$

(b) $200 R$

(d) $- 100 R$

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Answer:

Correct Answer: 17. (d)

Solution:

$Δ W_{1} + Δ U_{1} = Δ Q_{1} \dots (i)$

$Δ W_{2} + Δ U_{2} = Δ Q_{2} \dots (i i)$

$Δ Q_{1} + Δ Q_{2} = 0$

$∴ {(n C_{p} Δ T)}_{1} + {(n C_{p} Δ T)}_{2} = 0$

But $n_{1} = n_{2} = 2$

$∴ \frac{5}{2} R (T - 700) + \frac{7}{2} R (T - 400) = 0$

Solving, we get $T = 525 K$

Now, from Eqs. (i) and (ii), we get

$\begin{array}{cc} Δ W_{1} + Δ W_{2} = - Δ U_{1} - Δ U_{2} \\ as & Δ Q_{1} + Δ Q_{2} = 0 \\ ∴ & Δ W_{1} + Δ W_{2} = - [{(n C_{V} Δ T)}_{1} + {(n C_{V} Δ T)}_{2}] \\ = - [2 \times \frac{3}{2} R \times (525 - 700) + 2 \times \frac{5}{2} R \times (525 - 400)] \\ = - 100 R \end{array}$

Heat and Thermodynamics 6 Question 13

17. Now consider the partition to be free to move without friction so that the pressure of gases in both compartments is the same. Then total work done by the gases till the time they achieve equilibrium will be

Answer:

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Heat and Thermodynamics 6 Question 13

17. Now consider the partition to be free to move without friction so that the pressure of gases in both compartments is the same. Then total work done by the gases till the time they achieve equilibrium will be

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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