A gas tank contains helium gas. If the tank is cooled to one third of its origin

Question

A gas tank contains helium gas. If the tank is cooled to one third of its original temperature, and then enough additional helium is added until there is twice as much as before, how does the new pressure of the tank compare to the old pressure? A weather balloon is filled with helium at 25.0 degree C and 1.00 bar. At this point the volume of the balloon is 250.0L. The balloon rises to a point where T = -30.0 degree C and it has expanded to 850.0 L. What is the new pressure? Calculate the volume of 5.00g of nitrogen at STP.

Explanation / Answer

(14) Ideal gas equation,

P V = n R T
Here for the present condition, V and R are constant,

P1/ n1 T1 = P2 / n2 T2

1 / 1 * 1 = P2 / 2 * (1/3)

P2 = 2/3, So pressure is reduced to 2/3 rd of its original pressure.

(15)

Ideal gas equation, P V = n R T

Here, n and R are constant.

P1 V1 / T1 = P2 V2 / T2

1.00 * 250.0 / 298.15 = P2 * 850.0 / 243.15

P2 = 0.240 bar

(16)

Moles of N2 = mass / molar mass = 5.00 / 28.0 = 0.178 mol

At STP, 1 mol of any gas occupies 22.4 L of volume

then, 0.178 mol of N2 gas occupies 22.4 * 0.178 = 4.00 L of volume

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