Question

A closed Vessel contains air at a temperature 30 C and a pressure of one atm. it is heated to 150 C. What will be the new pressure??

Answer 1

Question #35559

A closed Vessel contains air at a temperature 30 C and a pressure of one atm. It is heated to 150 C. What will be the new pressure??

Solution:

Let:

$$t _ {1} = 3 0 {}^ {\circ} \mathrm {C}$$ $$t _ {2} = 1 5 0 {}^ {\circ} \mathrm {C}$$ $$P _ {1} = 1 \text{ atm}$$ $$P _ {2} = ?$$

According to the Gay Lussac's law (isochoric process)

$$\frac {P _ {1}}{T _ {1}} = \frac {P _ {2}}{T _ {2}}$$ $$P _ {2} = P _ {1} \frac {T _ {2}}{T _ {1}} \text{ were } T _ {1} = t _ {1} + 2 7 3 K, T _ {2} = t _ {2} + 2 7 3 K, \text{ K-Kelvin}$$ $$T _ {1} = 3 0 + 2 7 3 = 3 0 3 K$$ $$T _ {2} = 1 5 0 + 2 7 3 = 4 2 3 K$$ $$P _ {2} = 1 * \frac {4 2 3}{3 0 3} = 1.4 \text{ atm}.$$

Answer: 1.4 atm.