Question

A fixed amount of a particular ideal gas at 168C and a pressure of 1.75 3 105 Pa occupies a volume of 2.75 m3. If the volume is increased to 4.20 m3 and the temperature is raised to 26.48C, what will be the new pressure of the gas?

Answer 1

The new pressure of a gas is
`4.97 * 10^(5)` Pa

Step-by-step explanation:

Given:

Initial pressure
`P_(1) = 1.75 * 10^(5)` Pa

Initial temperature
`T_(1) = 16.8 + 273 = 289.8` K

Final temperature
`T_(2) = 26.48 + 273=299.48` K

Initial volume
`V_(1) = 2.75 m^(3)`

Final volume
`V_(2) = 4.20m^(3)`

From ideal gas equation,

`PV = nRT`

Where
`n =` number of moles, here
`n =1`,
`R =` gas constant

`(PV)/(T) = constant`

For finding new pressure of gas,

`(P_(1) V_(1) )/(P_(2) V_(2) ) =(T_(1) )/(T_(2) )`

`P_(2) = (P_(1) V_(1) T_(2) )/(T_(1) )`

`P_(2) = (1.75 * 10^(5) * 2.75 * 299.48)/(289.8)`

`P_(2) = 4.97 * 10^(5)` Pa

Therefore, the new pressure of a gas is
`4.97 * 10^(5)` Pa