Part A) A weather balloon is inflated to a volume of 25.0 L at a pressure of 755

Question

Part A)

A weather balloon is inflated to a volume of 25.0 L at a pressure of 755 mmHg and a temperature of 25.9 C. The balloon rises in the atmosphere to an altitude where the pressure is 375 mmHg and the temperature is -14.7 C. Assuming the balloon can freely expand, calculate the volume of the balloon at this altitude.

PartB)

At standard temperature and pressure (0 C and 1.00 atm ), 1.00 mol of an ideal gas occupies a volume of 22.4 L. What volume would the same amount of gas occupy at the same pressure and 50 C ?

Part C)

A balloon was filled to a volume of 2.60 L when the temperature was 30.0C. What would the volume become if the temperature dropped to 11.0C

Explanation / Answer

a)

V = 25

P = 755

T = 25.9 = 298.9

P2 = 375

T2 = -14.7 = 258.3

V2 = ?

Apply ideal gas law ratios

P1*V1/T1 = P2*V2/T2

V2 = P1/P2 * T2/T1 * V1

V2 = (755/375)(258.3/298.9) * 25 = 43.49 L

B)

Fin volume when T = 50 = 323 K

V1/T1 = V2/T2

V2 = T2/T1 * V1 = 323/273 * 22.4 = 26.50 L

C)

V = 2.6 L

T = 30 = 303

T = 11 = 284

V1/T1 = V2/T2

V2 = V1*T2/T1 = 2.6*284/303 = 2.43696 L

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