Question

For many purposes we can treat nitrogen N2 as an ideal gas at temperatures above its boiling point of −196.°C. Suppose the temperature of a sample of nitrogen gas is lowered from 18.0°C⁢ to −15.0°C, and at the same time the pressure is changed. If the initial pressure was 3.6atm and the volume decreased by 40.0%, what is the final pressure? Round your answer to the correct number of significant digits.

Answer 1

Answer:

5.3 atm

Explanation:

We can solve this problem using the equation of state for an ideal gas, which states that:

$pV=nRT$

where

p is the pressure of the gas

V is its volume

n is the number of moles

R is the gas constant

T is the absolute temperature of the gas

For a gas undergoing a transformation, n and R remain constant, so the equation can be written as

$\frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2}$

For the gas in this problem we have:

$T_1=18.0C+273=291 K$ is the initial temperature of the gas

$T_2=-15.0C + 273=258 K$ is the final temperature

$p_1=3.6 atm$ is the initial pressure

$V_2=0.60V_1$, as the volume is decreased by 40.0%

Solving for $p_2$, we find the final pressure of the gas:

$p_2 = \frac{p_1 V_1 T_2}{T_1 V_2}=\frac{(3.6)V_1(258)}{(291)(0.60V_1)}=5.3 atm$