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?

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,

[tex]\frac{V_1}{n_1}=\frac{V_2}{n_2}[/tex]

Final conditions are:

[tex]n_2=0.07mol[/tex]

[tex]V_2=?L[/tex]

Putting values in above equation, we get

[tex]\frac{0.06}{0.04}=\frac{V_2}{0.07}[/tex]

[tex]V_2=0.105L[/tex]

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,

[tex]\frac{V_1}{T_1}=\frac{V_2}{T_2}[/tex]

Final conditions are:

[tex]T_2=340K[/tex]

[tex]V_2=?L[/tex]

Putting values in above equation, we get

[tex]\frac{0.06}{240}=\frac{V_2}{340}[/tex]

[tex]V_2=0.085L[/tex]

mahima6824soni mahima6824soni · Answer 2 · 2022-03-03T13:35:40+01:00

The volume the gas will occupy if the number of moles is increased to 0.07 mol is 0.105L.

The volume the gas will occupy if the temperature is increased to 340.00 k (t2) is 0.085L.

Volume is the total space occupied by any object.

Given the pressure is 14.20 atm.

Volume = 0.06L

Temperature = 240 K

n = 0.04 mol

Increase number of moles are 0.07 mol

By Ideal gas law

[tex]\rm \dfrac{V_1}{n_1 } =\dfrac{V_2}{n_2}[/tex]

Putting the values

[tex]\rm \dfrac{0.06}{0.04 } =\dfrac{V_2}{0.07} = 0.105L[/tex]

Now, if the temperature is increased to 340.00 k

[tex]\rm \dfrac{V_1}{T_1 } =\dfrac{V_2}{T_2}\\\\\\\dfrac{0.06}{240 } =\dfrac{V_2}{340}= 0.085L[/tex]

Thus, volume 1 is 0.105 L and volume 2 is 0.085L.

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