Question

On a recent trip to the Fort Worth Zoo, Trevor's parents bought him a helium-filled balloon. At the beginning of the day, the sky was sunny and clear. The balloon was filled with 14.2 L of helium at a pressure of 102.5 kPa. The temperature was 33 °C. After a while, a storm came up, and a drop in barometric pressure and temperature reduced the pressure of the helium to 100.9 kPa and the temperature to 24 °C. What was the new volume of the balloon if there was a change? ______ Did the volume increase or decrease? _____

Answer 1

Answer:

1. The new volume is 14L

2. The volume decreased

Explanation:

Step 1:

Data obtained from the question.

Initial Volume (V1) = 14.2 L

Initial pressure (P1) = 102.5 kPa

Initial temperature (T1) = 33°C

Final pressure (P2) = 100.9 kPa

Final temperature (T2) = 24°C.

Final volume (V2) =?

Step 2:

Conversion of celsius temperature to Kelvin temperature. This is illustrated below:

K = °C + 273

T1 = 33°C = 33°C + 273 = 306K

T2 = 24°C = 24°C + 273 = 297K

Step 3:

Determination of the new volume.

Applying the general gas equation P1V1/T1 = P2V2/T2, the new volume can be obtain as follow:

P1V1/T1 = P2V2/T2

102.5 x 14.2/306 = 100.9 x V2/297

Cross multiply to express in linear form as shown below:

306 x 100.9 x V2 = 102.5 x14.2 x297

Divide both side by 306 x 100.9

V2 = (102.5x14.2x297)/(306 x 100.9)

V2 = 14 L

The new volume is 14 L.

Step 4:

Determination of the change in volume. This is illustrated below

Final volume (V2) = 14 L

Initial volume (V1) = 14.2 L

Change in volume (ΔV) =?

Change in volume (ΔV) = Final volume (V2) - Initial volume (V1)

ΔV = V2 - V1

ΔV = 14 - 14.2

ΔV = - 0.2L

Since the change in volume is negative, it means there is a decrease in the volume.

Answer 2

Answer:

The volume decreased to 14.0 L

Explanation:

Step 1: Data given

Volume of the balloon 14.2 L

Pressure = 102.5 kPa = 1.0115963 atm

Temperature = 33 °C = 306 K

The pressure reduced to 100.9 kPa = 0.99580558

Temperature reduced = 24.0 °C = 297 K

Step 2: Calculate the new volume

(P1*V1)/T1 = (P2*V2)/T2

⇒with P1 is the initial pressure = 102.5 kPa

⇒with V1 = the initial volume of the balloon = 14.2 L

⇒with T1 = the initial temperature = 306 K

⇒with P2 = the reduced pressure = 100.9 kPa

⇒with V2 = the final volume = TO BE DETERMINED

⇒with T2 = the reduced temperature = 297 K

(102.5 kPa * 14.2 L) . 306 K  = (100.9 kPa * V2) / 297 K

4.7565 = (100.9 *V2) / 297 K

1412.68 = 100.9 * V2

V2 = 14.00 L

The volume decreased to 14.0 L