Answer to Question #89308 in Molecular Physics | Thermodynamics for Shivam Nishad

Question

Two identical gas systems, each containing 0.1 mole of ideal gas, are at 300 K and 3 atm pressure. One system is allowed to expand adiabatically and the other isothermally, till both attain normal pressure. Show that the volume of adiabatically expanded
system is less than the isothermally expanded system. The ratio of the heat capacities of the gas is
1.4.

Answer 1

adiabatic equation

$$p_1V_1^{\gamma}=p_2V_2^{\gamma}$$

from this

$$(\frac{V_1}{V_2})^{1.4}=1/3$$ $$V_2=3^{\frac{1}{1.4}}V_1$$

isotherm equation

$$p_1V_1=p_2V_{2'}$$

from this

$$\frac{V_1}{V_{2'}}=\frac{1}{3}$$ $$V_{2'}=3V_1$$

In over

$$\frac{V_{2'}}{V_2}=3^{\frac{0.4}{1.4}}=1.37$$

$$V_{2'}=1.37\cdot V_2$$

So, in isotherm process volume 1.37 times more than volume adiabatic process