In a process, temperature|KTG & THERMODYNAMICS|JEEMAIN 2019

JEE MAIN 2019

11-01-2019 S2

Question

In a process, temperature and volume of one mole of an ideal monoatomic gas are varied according to the relation VT = K, where K is a constant. In this process, the temperature of the gas is increased by T. The amount of heat absorbed by gas is (R is gas constant):

Select the correct option:

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

$\frac{1}{2} R \Delta T$

$\frac{1}{2} \mathrm{KR} \Delta \mathrm{T}$

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

VT = K PV = RT ⇒ V·PV = K′ PV² = K′ ⇒ C = 3R/2 − R/(2 − 1) = R/2 ΔQ = nCΔT = 1/2 RΔT

Question Tags

JEE Main

Physics

Hard