Answer:
9,047.62 torr
Explanation:
The given parameters are pressure (P1) = 760 torr, Temperature at which the gas exerts pressure, P1, called the initial temperature (T1) = 21-celsius degrees, a second temperature(T2) = 250 Celsius degrees, and the pressure(P2) which such gas would exert at T2 is unknown [P2=?]. Notice that the temperature difference.
Gay Lussac's Law states that the pressure of a given amount of gas held at constant volume is directly proportional to the kelvin temperature.
If you heat a gas, the molecule gains kinetic energy and they move about in their container. As the temperature increase, the energy each particle gains would increase as well. Thus, allowing them to move faster and in all directions. Because they are confined to the boundary of the container, the gas particles bounce off the wall of the container. The energy allows continuous movement of the particle. Hence, the pressure each particle would exert on the wall of the container is equal to its momentum. Remember that the velocity, V, a particle is traveling with is promotional to the energy, E, that it obtains from the increase in temperature.
Using the Gay Lussac's Law equation: P1T2=P2T1 [Notice here that P=T].
Making P2 the mathematical subject, P2= P1T2/T1. ----Formula for the condition of this problem.
Substituting the known values;
P2= (760 torr x 250°C)/ 21 °C
P2= (190,000 °C torr)/21 °C
Cancel like terms in the numerator and denominator.
P2= (190,000 torr)/21
P2= 9,047.62 torr
[Notice that the pressure increase with the increase in temperature. The pressure increase from 760 torr to 9,047.62 torr with an increase in temperature from 21 Celsius degrees to 250 Celsius degrees.]
