V - k PT0-33 where k is constant- It is an-

The correct option is D isobaric processThe process in which change in #10;volume and temperature of a gas take place at a constant pressure is called an # ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Thermodynamic Processes

V = k PT0.33 ...

Question

$V = k {(\frac{P}{T})}^{0.33}$ where k is constant. It is an:

isothermal process

No worries! We‘ve got your back. Try BYJU‘S free classes today!

adiabatic process

No worries! We‘ve got your back. Try BYJU‘S free classes today!

isochoric process

No worries! We‘ve got your back. Try BYJU‘S free classes today!

isobaric process

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D isobaric process
The process in which change in volume and temperature of a gas take place at a constant pressure is called an isobaric process.
Consider some gas contained in a barrel. Supply some quantity of heat to increase its temperature from T1 to T2 without changing its pressure ‘P’. It will be observed that the gas expand. The change is said to be isobaric change

Suggest Corrections

Similar questions

V = k {(\frac{P}{T})}^{0.33}

where k is a constant, represents an

Q. For a given exothermic reaction,

K_{p}

and

K_{p}^{'}

are the equilibrium constants at temperatures

T_{1}

and

T_{2}

respectively. Assuming that heat of reaction is constant in temperatures range between

T_{1}

and

T_{2}

. It is readily observation that :
(Given,

T_{2} > T_{1}

)

Q. If stress-strain relation for volumetric change is in the form

\frac{Δ V}{V_{0}} = K P

where

P

is applied uniform pressure, then

K

stands for

Q. A particle is moving with a velocity

\to v = K (y^i + x^j),

where

K

is a constant. The general equation for its path is:

Q. Prove

a = \sqrt{\frac{K_{p}}{P + K_{p}}}

for

P C l_{5} ⇌ P C l_{3}

C l_{2}

, where

a

is degree of dissociation at temperature when equilibrium constant is

K_{p}

Join BYJU'S Learning Program

V=k(PT)0.33 where k is constant. It is an:

$V = k {(\frac{P}{T})}^{0.33}$ where k is constant. It is an: