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 = ((nR)/(V))T`

so we can say

`(P_1)/(P_2) = (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

`(5)/(P_2) = (254.5)/(101.8)`

by solving above equation we will have

`P_2 = 2 atm`

so final pressure will be 2 atm