To understand the ideal gas law and be able to apply it to a wide variety of situations. The absolute temperature T, volume V, and pressure p of a gas sample are related by the ideal gas law, which states that pV=nRT. Here n is the number of moles in the gas sample and R is a gas constant that applies to all gases. This empirical law describes gases well only if they are sufficiently dilute and at a sufficiently high temperature that they are not on the verge of condensing. In applying the ideal gas law, p must be the absolute pressure, measured with respect to vacuum and not with respect to atmospheric pressure, and T must be the absolute temperature, measured in kelvins (that is, with respect to absolute zero, defined throughout this tutorial as ^ -273˚C). If p is in pascals and V is in cubic meters, use R=8.3145J/(mol x K). If p is in atmospheres and V is in liters, use R=0.08206L x atm/(mol x K) instead. Part A A gas sample enclosed in a rigid metal container at room temperature (20.0˚C) has an absolute pressure p1. The container is immersed in hot water until it warms to 40.0˚C. What is the new absolute pressure p2?

Respuesta :

Answer: New absolute pressure is [tex]1.07\times p_1[/tex]

Explanation:

Gay-Lussac's Law:

This law states that pressure is directly proportional to the temperature of the gas at constant volume and number of moles.  As the gas is enclosed in a rigid metal container the volume of the gas is fixed.

[tex]P\propto T[/tex]     (At constant volume and number of moles)

[tex] \frac{P_1}{T_1}=\frac{P_2}{T_2} [/tex]

Given:  

[tex] P_1 =p_1[/tex]

[tex]T_1=20^0C=(20+273)=293 K[/tex]

[tex] P_2=p_2[/tex]    

[tex] T_2=40^0C=40+273=313 K[/tex]

[tex] \frac{p_1}{293}=\frac{P_2}{313} [/tex]

[tex]p_2=1.07\times p_1[/tex]