Question

Calculate the heat capacity of a gas sample from the following information: The sam- ple comes to equilibrium in a flask at 25°C and 121.3 kPa. A stopcock is opened briefly, allowing the pressure to drop to 101.3 kPa. With the stopcock closed, the flask warms, returning to 25°C, and the pressure is measured as 104.0 kPa. Determine CP in J·mol−1·K−1 assuming the gas to be ideal and the expansion of the gas remaining in the flask to be reversible and adiabatic.

Answer 1

Answer : The value of
`c_p` for reversible and adiabatic expansion is 55.04 J/mol.K

Explanation : Given,

Temperature at equilibrium =
`T_1=25^oC=273+25=298K`

Pressure at equilibrium =
`P_1=121.3kPa`

Temperature at adiabatic reversible expansion =
`T_2`

Pressure at adiabatic reversible expansion =
`P_2=101.3kPa`

Temperature at constant volume process =
`T_3=25^oC=273+25=298K`

Pressure at constant volume process =
`P_3=104.0kPa`

First we have to calculate the temperature at adiabatic reversible expansion.

Gay-Lussac's Law : It is defined as the pressure of the gas is directly proportional to the temperature of the gas at constant volume and number of moles.

`P\propto T`

or,

`(P_2)/(T_2)=(P_3)/(T_3)`

Now put all the given values in the above equation, we get:

`(101.3kPa)/(T_2)=(104.0kPa)/(298K)`

`T_2=290K`

Now we have to calculate the value of
`c_p` for reversible and adiabatic.

Formula used :

`T_2=T_1((P_2)/(P_1))^{(R)/(c_p)}`

Now put all the given values in the above equation, we get:

`290=298* ((101.3)/(121.3))^{(8.314)/(c_p)}`

`c_p=55.04J/mol.K`

Therefore, the value of
`c_p` for reversible and adiabatic expansion is 55.04 J/mol.K