Answer to Question #201439 in Molecular Physics | Thermodynamics for Cadmium John

Question

The temperature of an ideal gas in a sealed 0.40m³ rigid container is reduced from 350 K to 270 K.The final pressure of the gas is 60 kPa. The molar heat capacity at constant volume of the gas is 28.0 J/(mol K). The heat absorbed by the gas is closest to

• A. -24 kJ.
•  B. -31 kJ.
•  C. 24 kJ.
•  D. 31 kJ.
•  E. None of the above.

Answer 1

Since the chamber is rigid, we have a constant volume and once we have the final state parameters we can calculate the number of mole of gas using the ideal gas law:

$n=\frac{pV}{RT}=\frac{(60000\,Pa)(0.40\,m^3)}{(8.314\,J/molK)(270\,K)}=10.6915\,mol\,of\,gas$

Then, we substitute on the equation that relates heat with molar heat capacity, the temperature differences, and the number of moles to find the amount of heat when the volume is constant:

$Q = nC_V\Delta T = (10.6915\,mol)(28.0 \frac{J}{molK})(270-350)K$

$Q = nC_V\Delta T = -23948.96\,J*\frac{1\,kJ}{10^3\,J} = -23.95\,kJ$

In conclusion, the heat absorbed by the gas is closest to option A

• A. -24 kJ.

Reference:

• Chang, R. (2010). Chemistry. McGraw-Hill, New York.