JEE Main 2019 (Online) 11th January Morning Slot | Heat and Thermodynamics Question 392 | Physics

A rigid diatomic ideal gas undergoes an adiabatic process at room temperature. The relation between temperature and volume for this process is TV^x = constant, then x is :

$${5 \over 3}$$

$${2 \over 5}$$

$${3 \over 5}$$

$${2 \over 3}$$

JEE Main 2019 (Online) 11th January Morning Slot

MCQ (Single Correct Answer)

-1

A gas mixture consists of 3 moles of oxygen and 5 moles of argon at temperature T. considering only translational and rotational modes, the total internal energy of the system is :

12 RT

20 RT

4 RT

15 RT

JEE Main 2019 (Online) 10th January Evening Slot

MCQ (Single Correct Answer)

-1

Half mole of an ideal monoatomic gas is heated at constant pressure of 1 atm from 20^oC to 90^oC. Work done by gas is close to – (Gas constant R = 8.31 J/mol.K)

581 J

73 J

146 J

291 J

JEE Main 2019 (Online) 10th January Evening Slot

MCQ (Single Correct Answer)

-1

Two kg of a monoatomic gas is at a pressure of 4 $$ \times $$ 10⁴ N/m². The density of the gas is 8 kg/m³. What is the order of energy of the gas due to its thermal motion ?

10⁴ J

10³ J

10⁵ J

10⁶ J

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