For a process, relation between temperature and volume is `TV^(3)`=constant. If a monoatomic gas follows this process, find molar specific heat for this process

Show that in an adiabatic process the relation between volume and temperature of a gas is TV^(gamma-1) = constant, where gamma is the ratio of the two specific heat of the gas.

Show that in an adiabatic process the relation between volume and temperature of a gas is TV^(gamma-1) . Constant, where gamma is the ratio of the two specific heats of the gas.

For a gas undergoing an adiabatic process, the relation between temperature and volume is found to be TV^(@.4) = constant. The gas may be

One mole of a monoatomic ideal gas undergoes process AB in given P – V diagram then average specific heat for this process is:

One mole of a monoatomic ideal gas undergoes the process ArarrB in the given P-V diagram. What is the specific heat for this process?

One mole of a monoatomic ideal gas undergoes the process ArarrB in the given P-V diagram. What is the specific heat for this process?

A monoatomic gas undergoes a process in which the pressure (P) and the volume (V) of the gas are related as PV^(-3)= constant. What will be the molar heat capacity of gas for this process?

A monoatomic gas undergoes a process in which the pressure (P) and the volume (V) of the gas are related as PV^(-3)= constant. What will be the molar heat capacity of gas for this process?

Molar specific heat at constant volume is C for a monoatomic gas is