Question

A container id divided into two compartment. One compartment contains $2$ mole of $N_2$ gas at $1$ atm and $300K$ & other compartment contains $H_2$ gas at the same temperature and pressure. Volume of $H_2$ compartment is four times the volume of $N_2$ compartment. [assume no reaction under these condition]
If the container containing $N_2$ and $H_2$ are further heated to $1000K$ forming $N{H}_3$ with $100\%$ yield. Calculate the final total pressure.

• A. $2.22atm$
• B. $3atm$
• C. $2atm$
• D. $3.33atm$

Answer 1

Answer: (D)

$3.33atm$

Compartment containing $N_2$

$n=2$

$P=1$atm

$T=300$K

Finding for V

Using Ideal gas equation

$PV=nRT$

$V=\frac{2\times 0.0821\times 300}{1}=49.26$L

Compartment containing $H_2$

Givn volume is 4 times than $N_2$

$V_2=4\times 49.26=197$L

Finding moles for $H_2$ Compartment

given pressure and temperature is equal to $N_2$ compartment

$n=\frac{197}{0.0821\times 300}=8$

Calculating final pressure

$T=1000$K

$P=\frac{8\times 0.0821\times 1000}{197}=3.33$atm