Question

A gas is held at 3.8 atm and 500 K. If the pressure is then decreased to 1.2 atm, what will the new temperature be?

Answer 1

`T_2 = 157.89 K`

`T_2 = -115.10 ^0C`

Explanation:

By changing the physical conditions like pressure, temperature and volume of the gas, the moles of the gas remain constant.
`n_1=n_2`

According to the ideal gas equation,

`(P_1V_1)/(RT_1) = (P_2V_2)/(RT_2)`

The volume of the process is also constant . As, according to the Avogadro's law,
`V alpha n`

So, `V_1=V_2`

`(P_1)/(T_1) = (P_2)/(T_2)`

`3.8/500 = 1.2/ T_2`

`T_2 =1.2/3.8 xx500`

`T_2 = 157.89 K`

`T_2 = -115.10 ^0C`

So, the new temperature will be `2280 K`.