A sample of gas exerts a pressure of 16 atm at a temperature of 340 K. What is the pressure of the gas in atm if the temperature is raised to 408 K?
ANSWER FAST

P2 = 19.2atm

Explanation:

Initial pressure (P1) = 16atm

Initial temperature (T1) = 340K

Final temperature (T2) = 408K

Final pressure (P2) = ?

This question involves the use of pressure law

Pressure law states that the pressure of a fixed mass of gas is directly proportional to it's temperature provided that volume is kept constant.

Mathematically,

P = kT, k = P / T

Therefore,

P1 / T1 = P2 / T2 = P3 / T3 = =Pn / Tn

P1 / T1 = P2 / T2

We need to solve for P2

P2 = (P1 × T2) / T1

Now we can plug in the values and solve for P2

P2 = (16 × 408) / 340

P2 = 6528 / 340

P2 = 19.2atm

The final pressure (P2) of the gas is 19.2atm

it is stable as it is in group 0 which have full outer shells of electrons