Answer to Question #100863 in Molecular Physics | Thermodynamics for Pham Anh Khoa

Question

An ideal monatomic gas undergoes an adiabatic compression from state 1 with pressure p1=1 atm, volume V1=8 L, and temperature T1=300 K to state 2 with pressure p2=32 atm, volume V2=1 L. (a) What is the temperature of the gas in state 2? (b) How many moles of gas are present?
(c) What is the average translational kinetic energy per mole before and after the compression? (d) What is the ratio of the squares of the rms speeds before and after the compression? (e) If we do not know that the ideal gas here is monatomic, demonstrate that the gas is truly monatomic.

Answer 1

a) Find a temperature in state 2 using this equation

$$T_1V_1^{\gamma-1}=T_2V_2^{\gamma-1}$$ $$T_2=(\frac{V_1}{V_2})^{\gamma-1}T_1$$ $$T_2=(\frac{8}{1})^{\frac{5}{3}-1}\cdot 300=1200 (K)$$

b) moles of gas

$$\nu=\frac{RT_1}{P_1V_1}$$ $$\nu=\frac{8.31\cdot 300}{10^5 \cdot 0.008}=3 (moles)$$

c)The average translation kinetic energy per mole

before

$$<E_1>=\frac{3}{2}RT_1$$ $$<E_1>=\frac{3}{2}\cdot 8.31\cdot 300=3739,5 (\frac{J}{mol})$$

after

$$<E_2>=\frac{3}{2}\cdot 8.31\cdot 1200=14958(\frac{J}{mol})$$

d) The ratio of the squares of the rms speeds

$$\frac{m_0v^2}{2}=\frac{3}{2}kT$$ $$\frac{v_1^2}{v_2^2}=\frac{T_1}{T_2}=\frac{300}{1200}=0.25$$

e)Show that the gas is truly monoatomic

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

from this

$$\gamma=\frac{ln(\frac{p_2}{p_1})}{ln(\frac{V_1}{V_2})}=1.667=\frac{5}{3}$$

it is monoatomic gas