Chemistry - Thermodynamics - I Question with Solution | TestHub

Question Explanation:

The question asks about the efficiency of the cycle.

Concept:

This question is based on the work and efficiency of the cycle.

Solution:

The absorbed heat $q_{abs}$ is given by:

$$q_{abs}=n{C}_p\Delta T=1\times \frac{5R}{2}\times 300=750R$$

The total work done $W_{total}$ is:

$$W_{total}=W_I+W_{II}+W_{III}=-1\times R\times 300+0+1\times R\times 300\ln 2$$

Therefore, the magnitude of work $|W|$ is:

$$|W|=300R(1-0.7)=300R\times 0.3=90R$$

The efficiency is calculated as:

$$=\frac{|W|}{q_{abs}}\times 100=\frac{90}{750}\times 100=12\%$$

Final Answer:

The correct option is (D).