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,

$\frac{P_2}{T_2}=\frac{P_3}{T_3}$

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

$\frac{101.3kPa}{T_2}=\frac{104.0kPa}{298K}$

$T_2=290K$

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

Formula used :

$T_2=T_1(\frac{P_2}{P_1})^{\frac{R}{c_p}}$

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

$290=298\times (\frac{101.3}{121.3})^{\frac{8.314}{c_p}}$

$c_p=55.04J/mol.K$

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