A gas at a pressure of `5.0` atm is heated from `0^(@)C` to `546^(@)C` and simultaneously compressed to one-third of its original volume. Hence final pressure is

Options

A. `10.0 atm`
B. `30.0 atm`
C. `45.0 atm`
D. `5.0 atm`

Answer

C

Explanation

`P_(1)= 5.0 atm, T_(1)= 0^(@)C= 273, V_(1)=V`
`P_(2)=? T_(2)= 546^(@)C= 819 K, V_(2)= (1)/(3)V`
`:. From (P_(1)V_(1))/(P_(2)V_(2))=(T_(1))/(T_(2))`
`implies (5xxV)/(P_(2)xx(1)/(3)V)= (273)/(819)`
`implies (5xx3)/(P_(2))=(1)/(3)`
`P_(2)= 45 atm`