The volume of a certain mass of gas at `N.T.P.` is `27.5 cm^(3)`. What pressure will be necessary to keep the volume same at `27^(@)C`?

Options

A. `11.13` bar
B. `1.113 ` bar
C. `2.15` bar
D. `1.56` bar

Answer

B

Explanation

`V_(1) = 27.5 cm^(3) , V_(2) = 27.5 cm^(3)`
`P_(1) = 1.013 "bar" , P_(2) = ?`
`T_(1) = 273 K , T_(2) = 177 + 273`
`=300 K , = 450 K`
According to ideal gas equation,
`(P_(1)V_(1))/(T_(1)) = (P_(2)V_(2))/(T_(2))` or `V_(2) = (P_(1)V_(1)T_(2))/(T_(1)P_(2))`
`= ((1"bar") xx (VL) xx (450 K))/((300 K) xx (1.5 "bar")) = VL`