Question

A gas bottle contains 0.650 mol of gas at 730 mm Hg pressure. If the final pressure is 1.15 atm, how many moles of gas were added to the bottle?

Answer 1

Answer: 0.129 moles of gas were added to the bottle

Explanation:

According to the ideal gas equation:'

$PV=nRT$

P = Pressure of the gas

V= Volume of the gas

T= Temperature of the gas

R= Gas constant

n=  moles of gas

As Volume , gas constant and temperature are constant

$\frac{P_1}{n_1}=\frac{P_2}{n _2}$

where,

$P_1$ = initial pressure of gas =730 mm Hg =0.960 atm (760 mmHg = 1 atm)

$P_2$ = final pressure of gas = 1.15 atm

$n_1$ = initial number of moles = 0.650

$n_2$ = final number of moles =  ?

Now put all the given values in the above equation, we get the final moles of gas.

$\frac{0.960}{0.650}=\frac{1.15}{n_2}$

$n_2=0.779$

Therefore, the number of moles of gas will be 0.779

Thus moles of gas were added to the bottle are (0.779-0.650) = 0.129

Answer 2

Answer:

0.779 mol

Explanation:

Since the gas is in a bottle, the volume of the gas is constant. Assuming the temperature remains constant as well, then the gas pressure is proportional to the number of moles:

$p \propto n$

so we can write

$\frac{p_1}{n_1}=\frac{p_2}{n_2}$

where

p1 = 730 mm Hg = 0.96 atm is the initial pressure

n1 = 0.650 mol is the initial number of moles

p2 = 1.15 atm is the final pressure

n2 is the final number of moles

Solving for n2,

$n_2 = n_1 \frac{p_2}{p_1}=(0.650 mol)\frac{1.15 atm}{0.96 atm}=0.779 mol$