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 ^ -273ËšC). 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 (20.0ËšC) has an absolute pressure p1. The container is immersed in hot water until it warms to 40.0ËšC. 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]

good job