Physics, 12.11.2019 02:31 jackiecroce1

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.

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Answer from: amcda213040

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

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