The correct order of the values of critical temp- T-C- Boyle temp- T-B and inversion temp- T-i of a real gas is-

Question

Toppr Admin · Accepted Answer

Correct option is A- -T-C -T-C-dfrac-8-27- T-B-T-i- -dfrac-27-4-T-C-dfrac-T-B-T-C-dfrac-12-8-dfrac-Ti-T-C-dfrac-27-4-times -dfrac-2-2-dfrac-27-times 2-8-T-i- -gt- T-B -gt- T-C-Hence- the correct answer is option -A-

The correct order of the values of critical temp, $$T_C$$; Boyle temp, $$T_B$$ and inversion temp, $$T_i$$ of a real gas is:

Correct option is A. $$T_C < T_B < T_i$$
$$T_C=\dfrac{8}{27} T_B$$

$$T_i$$ $$=\dfrac{27}{4}T_C$$

$$\dfrac{T_B}{T_C}=\dfrac{12}{8}$$

$$\dfrac{Ti}{T_C}=\dfrac{27}{4}\times \dfrac{2}{2}=\dfrac{27\times 2}{8}$$

$$T_{i} > T_B > T_C$$

Hence, the correct answer is option $$A$$.

The correct order of the values of critical temp, $$T_C$$; Boyle temp, $$T_B$$ and inversion temp, $$T_i$$ of a real gas is:

Correct option is A. $$T_C < T_B < T_i$$$$T_C=\dfrac{8}{27} T_B$$$$T_i$$ $$=\dfrac{27}{4}T_C$$$$\dfrac{T_B}{T_C}=\dfrac{12}{8}$$$$\dfrac{Ti}{T_C}=\dfrac{27}{4}\times \dfrac{2}{2}=\dfrac{27\times 2}{8}$$$$T_{i} > T_B > T_C$$Hence, the correct answer is option $$A$$.

Correct option is A. $$T_C < T_B < T_i$$
$$T_C=\dfrac{8}{27} T_B$$

$$T_i$$ $$=\dfrac{27}{4}T_C$$

$$\dfrac{T_B}{T_C}=\dfrac{12}{8}$$

$$\dfrac{Ti}{T_C}=\dfrac{27}{4}\times \dfrac{2}{2}=\dfrac{27\times 2}{8}$$

$$T_{i} > T_B > T_C$$

Hence, the correct answer is option $$A$$.