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Question: In outer space, the pressure recorded is \(5 \times {10^{ - 4}}torr\). How much outer space could be...

In outer space, the pressure recorded is 5×104torr5 \times {10^{ - 4}}torr. How much outer space could be compressed into 1dm31d{m^3} box at a pressure of 1atm1atm?
A) 1.52×1016dm31.52 \times {10^{16}}d{m^3}
B) 4.56×1016dm34.56 \times {10^{16}}d{m^3}
C) 2.28×1016dm32.28 \times {10^{16}}d{m^3}
D) 1.14×1016dm31.14 \times {10^{16}}d{m^3}

Explanation

Solution

We know that Boyle’s Law. The volume of the gas is inversely related at constant temperature and pressure.
PV = K{\text{PV = K}}
Where,
P is pressure of the gas at constant temperature.
V is volume of the gas at constant temperature.
K is the constant.

Complete step by step answer:
The volume or pressure of the gas can be calculated using the relation.
P1V1 = P2V2{{\text{P}}_{\text{1}}}{{\text{V}}_{\text{1}}}{\text{ = }}{{\text{P}}_{\text{2}}}{{\text{V}}_{\text{2}}}
Where {{\text{P}}_{\text{1}}}{\text{ & }}{{\text{V}}_{\text{1}}} are pressure and volume of gas at initial state.
The pressure and volume of gas at final state is {{\text{P}}_{\text{2}}}{\text{ & }}{{\text{V}}_{\text{2}}}
Given,
The pressure at initial state (P1)\left( {{{\text{P}}_{\text{1}}}} \right) is 5×104torr(6.57×107atm)5 \times {10^{ - 4}}torr(6.57 \times {10^{ - 7}}atm)
The volume at initial state (V1)\left( {{{\text{V}}_{\text{1}}}} \right) is 1dm3{\text{1}}d{m^3}.
The pressure at final state (P2)\left( {{{\text{P}}_{\text{2}}}} \right) is 1atm{\text{1atm}}.
The volume at final state (V2)\left( {{{\text{V}}_{\text{2}}}} \right) can be calculated as,
P1V1 = P2V2{{\text{P}}_{\text{1}}}{{\text{V}}_{\text{1}}}{\text{ = }}{{\text{P}}_{\text{2}}}{{\text{V}}_{\text{2}}}
V2 = P1V1P2{{\text{V}}_{\text{2}}}{\text{ = }}\dfrac{{{{\text{P}}_{\text{1}}}{{\text{V}}_{\text{1}}}}}{{{{\text{P}}_{\text{2}}}}}
Substituting the known values in formula we get,
V2 = (1atm)(1dm3)(6.57×107atm)\Rightarrow {{\text{V}}_{\text{2}}}{\text{ = }}\dfrac{{\left( {{\text{1atm}}} \right)\left( {1d{m^3}} \right)}}{{\left( {6.57 \times {{10}^{ - 7}}atm} \right)}}
On simplifying we get,
V2 = 1.52×1016dm3\Rightarrow {{\text{V}}_{\text{2}}}{\text{ = }}1.52 \times {10^{16}}d{m^3}
Therefore option A is correct.

Note:
As we know that the Boyles' law explains the reason why the bubble at the bottom of a glass of a soft drink gets larger when it rises to the surface and why the bag of potato chips on a plane ride appeared to be inflated when taken out,
According to Boyle's law, as the pressure decreases the volume increases thus, the bubbles at the bottom of a glass of a soft drink get larger when it rises to the surface because when the bubbles rise, the pressure inside the bubble starts to get reduced and the volume (size) of the bubble increases.
The bag of potato chips on a plane ride appeared to be inflated when taken out because according to Boyle's law whenever the pressure of a gaseous system decreases its volume increases at a constant temperature. At the higher height, the pressure is reduced so the gas inside the bag becomes inflated.