Answer to Question #197502 in Software Engineering for archie

Question #197502

A balloon has a volume of 4.0 L when at sea level (1.0 atm) at room temperature 0f 28 . C What will be its volume when inflated with inflated with the same amount of gas at an elevation where the atmospheric pressure is 700 mm Hg at 28 . C


1
Expert's answer
2021-05-24T10:14:37-0400

The ideal-gas equation


"PV=nRT"


The amount of the gas is the same.

Then


"\\dfrac{P_1V_1}{T_1}=\\dfrac{P_2V_2}{T_2}=\\text{constant}"

The balloon has a volume of 4.0 L when at sea level (1.0 atm) at a room temperature of 28 °C.

"V_1=4.0\\ L=4\\times 10^{-3}\\ m^3,"

"P_1=1.0\\ atm=101325\\ Pa,"

"T_1=28\\degree C=(28+273)\\ K=301\\ K"


"1 mm\\ Hg = 133.322\\ Pa"

"P_2=700mm\\ Hg=(700\\cdot 133.322)\\ Pa=93325.4\\ Pa"

 "T_2=28\\degree C=(28+273)\\ K=301\\ K=T_1"          



"\\dfrac{P_1V_1}{T_1}=\\dfrac{P_2V_2}{T_2}"

Solve for "V_2"


"V_2=\\dfrac{P_1T_1}{P_2T_2}V_1"

Substitute

    

"V_2=\\dfrac{101325\\ Pa(301\\ K)}{93325.4\\ Pa(301\\ K)}\\cdot4\\times 10^{-3}\\ m^3"

"\\approx 4.343\\times 10^{-3}\\ m^3\\approx4.343\\ L"

The volume of gas will be "V_2=4.343\\ L."



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