Question

An ideal gas is taken through a cyclic thermodynamic process through four steps. The amounts of heat given to the system in these steps are Q{1} = 1000J, Q{2} = - 800J, Q{3} = 450J , Q{4} = - 200 J respectively. The efficiency of cycle is nearly

Need full Explanation ! Answer is already marked in attachment just need the explanation or the way to reach the answer​

Answer 1

Hello There!!

We are given to find efficiency of the cycle.

We Know

Efficiency is denoted by the symbol "η"

As it is a cyclic process, So,

`\triangle \text{U} = 0 \\ \implies \triangle \text{ Q} = \triangle \text{W} + \triangle \text {U} \\ \implies\triangle \text{ Q} = \triangle \text{W} + 0 \\ \implies \triangle \text{Q} = \triangle \text{W}`

`\text{Let us first calculate} \: \triangle \text{Q}`

`\triangle \text{Q} = \text{Q}_1 + \text{Q}_2 + \text{Q}_3 + \text{Q}_4`

`\triangle \text{Q} = 1000 + ( - 800) + 450 + (- 200) \\ \\ \implies\triangle \text{Q} = 1000 - 800 + 450 - 200 \\ \\ \implies \triangle \text{Q} = 450 \text{J}`

Now, We need to find the Heat Input.

We know, Heat Input is always the sum of all positive heats

`\because {∆Q_(in)} = \text{∆Q}_1 + \text{∆Q}_3 \\ \\ 1000 + 450 = 1450 \text{J}`

Now We Have To Find The η%

`\text{η\%} = \frac{ \text{450}}{ \text{1450 } } * 100 \\ \\ \implies \text{η\%} = \frac{ \text{450}}{ \text{145} \cancel{0 } } * 10 \cancel0 \\ \\ \implies\text{η\%} = \frac{ \text{450}}{ \text{145 } } * 10 \\ \\ \implies{ \text{η\%}} = \frac { \cancel{450}}{ \cancel{145}} * 10 \: \: \: \: \fbox{cancelling by 5} \\ \\ \implies{ \text{η\%}} = (90)/(29) * 10 \\ \\ \implies{ \text{η\%}} = 31.0344\% \\ \\ \implies{ \text{η\%}} = \red{31\%}`

Option A= 31% is the correct answer

Hope this helps!!