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?

Question

Emery Hancock

Chemistry

9 October, 14:15

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?

+3

Answers (1)

Know the Answer?

Melina · Answer 1

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.