The specific heat of the mixture of two gases at constant volume is `(13)/(6) R`. The ratio of the number of moles of the first gas to the second gas is 1:2.
The respective gases may be

If the specific heat of a gas at constant volume is 3/2R , then the value of gamma will be

The molar specific heat at constant volume of gas mixture is (13R)/(6) . The gas mixture consists of

The molar specific heat at constant volume of gas mixture is (13R)/(6) . The gas mixture consists of

In a mixture of gases, the volume content of a gas is 0.06% at STP. Calculate the number of molecules of the gas in 1 L of the mixture.

In a mixture of gases, the volume content of a gas is 0.06% at STP. Calculate the number of molecules of the gas in 1 L of the mixture.

The molar specific heat of a gas at constant volume is 12307.69 J kg^(-1) K^(-1) . If the ratio of the two specific heats is 1.65, calculate the difference between the two molar specific heats of gas.

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))

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))

Find the molar specific heat of mixture at constnat volume. It one mole of a monoatomic gas is mixed with three moles of a diatomic gas.