A balloon is filled with an ideal gas at room temperature at a pressure of 200 k

Question

A balloon is filled with an ideal gas at room temperature at a pressure of 200 kPa

a.) What would be the volume of the gas if the pressure remained constant and the temperature increased be 20 percent?
b.) What would be the temperature inside the balloon if the volume remained constant but the pressure were increased by 15 percent?
c.) What would be the pressure inside the balloon if the temperature remained constant but the volume was decreased by 10 percent?

This is all the infromation the book gives in asking the question I would like help in setting up the problem and show me the steps taking to reach the solution thank you.

Explanation / Answer

Ideal gas equation is PV = nRT

where P is pressure, V is volume, n is number of moles of gas, R is molar gas constant, T is temperature in K.

Since n and R are constant in the balloon, PV/T = constant.

So P1V1/T1 = P2V2/T2, where 1 and 2 represent the initial and final states of the balloon.

(a) P1 = P2 = P, V1 = V, V2 =?, T1 = T, T2 = 1.2T (20% increase)

P1V1/T1 = P2V2/T2

P x V/T = P x V2/1.2T

V2 = 1.2V = 1.2V1

Thus the volume is 1.2 times larger, i.e. it increases by 20%.

To find the final volume V2, you will need to know the initial volume V1, which is not possible to determine with the information provided. You will need to know either V1 or n to solve this part.

(b) P1 = P, P2 = 1.15P (15% increase), V1 = V2 = V, T1 = T, T2 = ?

P1V1/T1 = P2V2/T2

P x V/T = 1.15P x V/T2

T2 = 1.15T = 1.15T1

Thus the temperature is 1.15 times larger, i.e. it increases by 15%.

T1 = room temperature = 298 K

T2 = 1.15 x 298 = 343 K

(c) P1 = P, P2 = ?, V1 = V, V2 = 0.9V (10% decrease), T1 = T2 = T

P1V1/T1 = P2V2/T2

P x V/T = P2 x 0.9V/T

P2 = 1.11P = 1.11P1

Thus the pressure is 1.11 times larger, i.e. it increases by 11%.

P1 = 200 kPa

P2 = 1.11 x 200 = 222 kPa

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