Question

An open container of volume 5L contains air at 27°C and 1 atm. The container is heated to TK. The
amount of air expelled from the container is found to be 1.5L at-33ºC and 1 atm. What is the value of
temperature TK?​

Answer 1

Answer : The value of temperature TK is, 479.9 K

Explanation :

First we have to calculate the initial moles of air.

Using ideal gas equation,

$PV=nRT$

where,

P = pressure of air = 1 atm

V = volume of air = 5 L

T = temperature of air = $27^oC=273+27=300K$

n = number of moles of air = ?

R = gas constant = 0.0821 L.atm/mole.K

Now put all the given values in the ideal gas equation, we get

$(1atm)\times (5L)=n\times (0.0821L.atm/mole.K)\times (300K)$

$n=0.203mole$

Now we have to calculate the final moles of air.

Using ideal gas equation,

$PV=nRT$

where,

P = pressure of air = 1 atm

V = volume of air = 1.5 L

T = temperature of air = $-33^oC=273+(-33)=240K$

n = number of moles of air = ?

R = gas constant = 0.0821 L.atm/mole.K

Now put all the given values in the ideal gas equation, we get

$(1atm)\times (1.5L)=n\times (0.0821L.atm/mole.K)\times (240K)$

$n=0.0761mole$

Now we have to calculate the number of moles of air expelled from the container.

Number of moles of air expelled = Final moles of air - Initial moles of air

Number of moles of air expelled = 0.203 - 0.0761 = 0.1269 mole

Now we have to calculate the value of temperature TK.

Using ideal gas equation,

$PV=nRT$

where,

P = pressure of air = 1 atm

V = volume of air = 5 L

T = temperature of air = ?

n = number of moles of air = 0.1269 mole

R = gas constant = 0.0821 L.atm/mole.K

Now put all the given values in the ideal gas equation, we get

$(1atm)\times (5L)=(0.169mole)\times (0.0821L.atm/mole.K)\times T$

$T=479.9K$

Therefore, the value of temperature TK is, 479.9 K