Question

An ideal, monotomic gas initially at a temperature of 450K, a pressure of 4.00 atm and a volume of 10.0L, undergoes an adiabatic compression to 1/3 its original volume. Find the final temperature of the gas. A. 72 K B. 150 K C. 216 K D. 936 K E. 1350 K

Answer 1

Answer: D) 936 K

Step-by-step explanation:

Given:

Initial temperature of the gas,
`T = 450\ K`

Initial Pressure of the gas,
`P=4\ atm`

initial volume of the gas,
`V=10\ L`

It it given that the process is adiabatic, so for a adiabatic process we have

Let
`T_f \ \ and\ \ V_f` be the final temperature and volume of the gas.

`T_iV_i^(\gamma -1)=T_fV_f^(\gamma -1)`

For monotomic gas
`\gamma=1.67`

`450* V_i^(1.67 -1) =T_f\left ((V_i)/(3) \right )^(1.67-1)\\T_f=936 K`

Hence the final temperature of the gas is 936 K. So option D is correct