Question

The volume of an ideal gas is adiabatically reduced from 151 L to 80.6 L. The initial pressure and temperature are 1.50 atm and 260 K. The final pressure is 3.61 atm. (a) Is the gas monatomic, diatomic, or polyatomic? (b) What is the final temperature? (c) How many moles are in the gas?

Answer 1

Answer:

gas is dioatomic

T_f = 330.0 K

$\eta = 7.07 mole$

Explanation:

Part 1

below equation is used to determine the type Gas by determining $\gamma$ value

$\frac{V_{1}}{V_{F}}\gamma=\frac{P_{i}}{P_{f}}$

where V_i and V_f is initial and final volume respectively

and P_i and P_f are initial and final pressure

$\gamma = \frac{ln(P_f/P_i)}{ln(V_i/V_f)}$

$\gamma = \frac{ln(3.61/1.50)}{ln(151/80.6}$

\gamma = 1.38

therefore gas is dioatomic

Part 2

final temperature in adiabatic process is given as

$T_f = T_i*[\frac{v_i}{V_f}](^\gamma-1)$

substituing value to get final temperature

$T_f = 260*[\frac{151}{80.6}]^ {(1.38-1)}$

T_f = 330.0 K

Part 3

determine number of moles by using following formula

$\eta =\frac{PV}{RT}$

$\eta =\frac{1.013*10^{5}*0.151}{8.314*260}$

$\eta = 7.07 mole$