Question

A sample of a gas in a rigid container has an initial pressure of 5 atm at a temperature of 254.5 K. The temperature is decreased to 101.8 K. What is the new pressure?

Pls help! I give 8 points!

Answer 1

here since container is rigid so we can say here volume must be constant

so here we can say by ideal gas equation

PV = nRT

here we know that

P = pressure

V = volume

n = number of moles

R = gas constant

T = temperature

so here if volume is given constant

$P = (\frac{nR}{V})T$

so we can say

$\frac{P_1}{P_2} = \frac{T_1}{T_2}$

now we know that

$P_1 = 5 atm$

$T_1 = 254.5 K$

$T_2 = 101.8 K$

now from above equation we will have

$\frac{5}{P_2} = \frac{254.5}{101.8}$

by solving above equation we will have

$P_2 = 2 atm$

so final pressure will be 2 atm