JEE PHYSICS MCQ: THERMODYNAMICS

1 Medium

1 of 1

Application +4 | -1

5 mole of an ideal gas with

(\gamma = 7/5)

initially at STP are compressed adiabatically so that its temperature becomes

{400^ \circ }C

.The increase in the internal energy of gas in kJ is

View answer and explanation

Correct option: B

41.55

Option B (41.55) is correct because the increase in internal energy is calculated using the formula ∆U = nCv∆T, with Cv derived from γ, resulting in approximately 41.55 kJ.

More Questions from THERMODYNAMICS

Practice All Chapter MCQs →

The work done in which of the following process is zero? Water is used to cool the radiators of engines in cars because A gas expand

0.25\,\,{m^3}

at constant pressure

{10^3}N/{m^2},

the work done is A thermodynamic process in which temperature

T

of the system remains constant though other variable

P

and

V

may change, is called What is the slope for an isothermal process? For adiabatic processes

\left( {\gamma = \frac{{{C_p}}}{{{C_v}}}} \right)