A container has a volume of #3 L# and holds #2 mol# of gas. If the container is expanded such that its new volume is #8 L#, how many moles of gas must be injected into the container to maintain a constant temperature and pressure?

2 Answers
Jul 25, 2018

Answer:

The number of moles to be injected is #=3.33mol#

Explanation:

Apply Avogadro's Law

#"Volume"/"Number of moles"= " Constant"#

At constant temperature and pressure

#V_1/n_1=V_2/n_2#

The initial volume is #=3L#

The initial number of moles is #n_1=2mol#

The final volume is #V_2=8L#

The final number of moles is

#n_2=V_2/V_1*n_1=8/3*2=5.33mol#

The number of moles to be injected is

#=n_2-n_1=5.33-2=3.33mol#

Answer:

#10/3 \ \text{moles#

Explanation:

Assume the gas is ideal

Let #P_1# & #T_1# be the initial pressure & temperature of given gas (assumed ideal) having initial volume #V_1=3\ L# & moles #n_1=2 \ \text{moles#

Now, from ideal gas equation,

#PV=nRT#

#\implies P_1V_1=n_1RT_1#

#P_1(3)=2RT_1\ .........(1)#

Let #x\ \text{moles# of same gas be injected to the container such that final volume is #V_2=8\ L# & final moles in container become #n_2=2+x# such that the final temperature & pressure remain same t.e. #T_2=T_1# & #P_2=P_1#

Now, from ideal gas equation,

# P_2V_2=n_2RT_2#

#P_1(8)=(x+2)RT_1\ .........(2)#

Dividing (2) by (1) as follows

#\frac{P_1(8)}{P_1(3)}=\frac{(x+2)RT_1}{2RT_1}#

#8/3=\frac{x+2}{2}#

#x+2=16/3#

#x=16/3-2#

#x=10/3#

Thus, #10/3 \ \text{moles# can be added to the container to maintain same temperature & pressure