Question

Given a fixed amount of gas in a rigid container (no change in volume), what pressure will the gas exert if the pressure is initially 1.50 atm at 22.0oC, and the temperature is changed to 11.0oC?
A. 301 atm
B. 1.56 atm
C. 0.750 atm
D. 1.44 atm
E. 3.00 atm

Answer 1

Answer:

The pressure the gas will have if the pressure is initially 1.50 atm at 22.0 ° C and the temperature changes at 11.0 ° C is 1.44 atm (option D)

Explanation:

Gay Lussac's law indicates that, as long as the volume of the container containing the gas is constant, as the temperature increases, the gas molecules move more rapidly. Then the number of collisions against the walls increases, that is, the pressure increases. That is, the gas pressure is directly proportional to its temperature.

Gay-Lussac's law can be expressed mathematically as follows:

$\frac{P}{T}=k$

Where P = pressure, T = temperature, K = Constant

You have a gas that is at a pressure P1 and at a temperature T1. When the temperature varies to a new T2 value, then the pressure will change to P2, and then:

$\frac{P1}{T1}=\frac{P2}{T2}$

In this case:

• P1= 1.50 atm
• T1= 22 °C= 295 °K (being 0°C= 273 °K)
• P2= ?
• T2= 11 °C= 284 K

Replacing:

$\frac{1.5 atm}{295 K}=\frac{P2}{284 K}$

Solving:

$P2= 284 K*\frac{1.5 atm}{295 K}$

P2=1.44 atm

The pressure the gas will have if the pressure is initially 1.50 atm at 22.0 ° C and the temperature changes at 11.0 ° C is 1.44 atm (option D)