SATHEE: Heat and Thermodynamics 5 Question 7

Heat and Thermodynamics 5 Question 7

7. For the given cyclic process $C A B$ as shown for a gas, the work done is

(2019 Main, 12 Jan I)

(a) $5 J$

(b) $10 J$

(d) $30 J$

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Answer:

Correct Answer: 7. (b)

Solution:

Key Idea

In a cyclic thermodynamic process work done $=$ area under $p - V$ diagram. Also in clockwise cycle, work done is positive.

In the given cyclic process, work done $= \oint p d V =$ area enclosed by the cycle $= \frac{1}{2} \times$ base $\times$ height of triangle $(C A B)$ made by cycle

$= \frac{1}{2} (V_{2} - V_{1}) (p_{2} - p_{1})$

From graph, given

$\begin{aligned} V_{2} & = 5 m^{3}, V_{1} = 1 m^{3} \\ p_{2} & = 6 P a, p_{1} = 1 P a \\ = \frac{1}{2} (5 - 1) (6 - 1) = \frac{1}{2} \times 4 \times 5 = 10 J \end{aligned}$

Heat and Thermodynamics 5 Question 7

7. For the given cyclic process $C A B$ as shown for a gas, the work done is

Answer:

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Heat and Thermodynamics 5 Question 7

7. For the given cyclic process CAB as shown for a gas, the work done is

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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7. For the given cyclic process $C A B$ as shown for a gas, the work done is