Question

For this exercise, you can simulate the described conditions by changing the values in the run experiment tool of the simulation. to be able to measure the effects on the gas volume, slide the rspd bar in the properties box to v (l) so that the volume bar turns yellow. this way, volume becomes the only dependent variable. note that when volume is selected with rspd, its value can no longer be directly controlled. suppose a piston automatically adjusts to maintain a gas at a constant pressure of 14.20 atm . for the initial conditions, there are 0.04 mol of helium at a temperature of 240.00 k . this gas occupies a volume of 0.06 l under those conditions. what volume will the gas occupy if the number of moles is increased to 0.07 mol (n2) from the initial conditions? what volume will the gas occupy if the temperature is increased to 340.00 k (t2) from the initial conditions?

Answer 1

Answer: The volume of the gas when number of moles are increased is 0.105L.

And the volume of the gas when temperature is increased is 0.085L

Explanation: The pressure of the gas is held constant which is 14.20 atm

We are given Initial conditions, which are:

V = 0.06L

T = 240 K

n = 0.04 mol

• When the number of moles of gas are increased, we will use Avogadro's Law, which says that the volume is directly proportional to the number of moles of gas at constant temperature and pressure.

Mathematically,

$\frac{V_1}{n_1}=\frac{V_2}{n_2}$

Final conditions are:

$n_2=0.07mol$

$V_2=?L$

Putting values in above equation, we get

$\frac{0.06}{0.04}=\frac{V_2}{0.07}$

$V_2=0.105L$

• When the temperature of the gas is increased, we will use Charles's Law, which says that the volume is directly proportional to the temperature of the gas at constant pressure and number of moles.

Mathematically,

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

Final conditions are:

$T_2=340K$

$V_2=?L$

Putting values in above equation, we get

$\frac{0.06}{240}=\frac{V_2}{340}$

$V_2=0.085L$

Answer 2

1) ideal gas law: p·V = n·R·T.
p - pressure of gas.
V -volume of gas.
n - amount of substance.
R - universal gas constant.
T - temperature of gas.
n₁ = 0,04 mol, V₁ = 0,06 l.
n₂ = 0,07 mol, V₂ = 0,06 · 0,07 ÷ 0,04 = 0,105 l.
2) V₁ = 0,06 l, T₁ = 240,00 K.
T₂ = 340,00 K, V₂ = 340 · 0,06 ÷ 240 = 0,05 l.