Home
Class 11
CHEMISTRY
C(v) values for monoatomic and diatomic ...

`C_(v)` values for monoatomic and diatomic gases respectively are

A

`(1)/(2)R, (3)/(2)R`

B

`(3)/(2)R,(5)/(2)R`

C

`(5)/(2)R,(7)/(2)R`

D

`(3)/(2)R,(3)/(2)R`

Text Solution

Verified by Experts

The correct Answer is:
B

`C_(V)=(3)/(2)R,C_(P)-C_(V)=R`
Promotional Banner

Similar Questions

Explore conceptually related problems

Two ideal monoatomic and diatomic gases are mixed with one another to form an ideal gas mixture. The equation of the adiabatic process of the mixture is PV^(gamma)= constant, where gamma=(11)/(7) If n_(1) and n_(2) are the number of moles of the monoatomic and diatomic gases in the mixture respectively, find the ratio (n_(1))/(n_(2))

If one mole each of a monoatomic and diatomic gases are mixed at low temperature then C_(p)//C_(v) ratio for the mixture is :

1mole of triatomic 2 moles of monoatomic and 2 moles of diatomic gases are mixed in a container Find (C_(P))/(C_(V)) for the mixture.

One mole of an ideal monoatomic gas is mixed with one mole of an equimolar mixture of monoatomic and diatomic ideal gases. Find the value of lambda= (C_P /C_v) for the final mixture

Find the ratio of number of moles of a monoatomic and a diatomic gas whose mixture has a value of adiabatic exponent gammaa =3/2.

What will be the ratio of number of molecules of a monoatomic and a diatomic gas in a vessel, if the ratio of their partial pressures is 5 : 3 ?

If at same temperature and pressure, the densities for two diatomic gases are respectively d_(1) and d_(2) , then the ratio of velocities of sound in these gases will be

Assertion: The Poisson's ratio for diatomic gases is more than for monoatomic gases. Reason: Diatomic gases possess more degree of freedom.

1 mole of monoatomic and one mole of diatomic gas are mixed together. The value of C_(upsilon) is

A mixture of n_(1) moles of monoatomic gas and n_(2) moles of diatomic gas has (C_(p))/(C_(V))=gamma=1.5