JEE Main 2023 (Online) 25th January Evening Shift | Heat and Thermodynamics Question 81 | Physics

According to law of equipartition of energy the molar specific heat of a diatomic gas at constant volume where the molecule has one additional vibrational mode is :-

$$\frac{9}{2}R$$

$$\frac{5}{2}R$$

$$\frac{3}{2}R$$

$$\frac{7}{2}R$$

JEE Main 2023 (Online) 25th January Morning Shift

MCQ (Single Correct Answer)

-1

The root mean square velocity of molecules of gas is

Proportional to temperature ($$T$$)

Inversely proportional to square root of temperature $$\left( {\sqrt {{1 \over T}} } \right)$$

Proportional to square of temperature ($$T^2$$)

Proportional to square root of temperature ($$\sqrt T$$)

JEE Main 2023 (Online) 25th January Morning Shift

MCQ (Single Correct Answer)

-1

Out of Syllabus

A Carnot engine with efficiency 50% takes heat from a source at 600 K. In order to increase the efficiency to 70%, keeping the temperature of sink same, the new temperature of the source will be :

300 K

1000 K

900 K

360 K

JEE Main 2023 (Online) 24th January Evening Shift

MCQ (Single Correct Answer)

-1

Let $$\gamma_1$$ be the ratio of molar specific heat at constant pressure and molar specific heat at constant volume of a monoatomic gas and $$\gamma_2$$ be the similar ratio of diatomic gas. Considering the diatomic gas molecule as a rigid rotator, the ratio, $$\frac{\gamma_1}{\gamma_2}$$ is :

$$\frac{35}{27}$$

$$\frac{25}{21}$$

$$\frac{21}{25}$$

$$\frac{27}{35}$$

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