Question #344858

What volume did a helium-filled balloon have at 26.8 °C and 1.87 atm if its new volume was 63.2 mL at 2.22 atm and 9.6°C? O a. 79.6 O b. 209 O c. 70.7 O d. 56.5


1
Expert's answer
2022-05-26T18:49:03-0400

In an isolated system, the mole number remains the same:

n1=n2n_1=n_2


In the ideal gas law, the mole number is related to the temperature T, volume V, pressure P and gas constant R:

pV=nRTpV=nRT

n=pVRTn=\frac{pV}{RT}


Thus:

p1V1RT1=p2V2RT2\frac{p_1V_1}{RT_1} = \frac{p_2V_2}{RT_2}


The initial volume V1 can be deduced from the last equation:

V1=p2V2T1T2p1=2.22×63.2×26.89.6×1.87=209mlV_1 = \frac{p_2V_2T_1}{T_2p_1} = \frac{2.22 \times 63.2 \times 26.8}{9.6 \times 1.87} = 209 ml

Answer: b 209 ml


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS