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Sagot :
- V1=1.5L
- V2=0.92L
- P1=2.8atm
- P2=?
Using boyles law
[tex]\boxed{\sf v\propto \dfrac{1}{p}}[/tex]
[tex]\\ \sf\longmapsto P_1V_1=P_2V_2[/tex]
[tex]\\ \sf\longmapsto P_2=\dfrac{P_1V_1}{V_2}[/tex]
[tex]\\ \sf\longmapsto P_2=\dfrac{2.8\times 1.5}{0.92}[/tex]
[tex]\\ \sf\longmapsto P_2=\dfrac{4.2}{0.92}[/tex]
[tex]\\ \sf\longmapsto P_2=4.56atm[/tex]
[tex]\\ \sf\longmapsto P_2\approx 4.6atm[/tex]
Answer:
[tex]\boxed {\boxed {\sf 4.6 \ atm}}[/tex]
Explanation:
We are asked to find the new pressure given a change in volume. We will use Boyle's Law, which states the volume of a gas is inversely proportional to the pressure. The formula for this law is:
[tex]P_1V_1= P_2V_2[/tex]
Initially, the gas occupies 1.5 liters at a pressure of 2.8 atmospheres.
[tex]1.5 \ L * 2.8 \ atm = P_2V_2[/tex]
The volume is changed to 0.92 liters, but the pressure is unknown.
[tex]1.5 \ L * 2.8 \ atm = P_2* 0.92 \ L[/tex]
We are solving for the final pressure, so we must isolate the variable P₂. It is being multiplied by 0.92 liters. The inverse operation of multiplication is division, so we divide both sides by 0.92 L.
[tex]\frac {1.5 \ L * 2.8 \ atm}{0.92 \ L} = \frac{P_2* 0.92 \ L}{0.92 \ L}[/tex]
[tex]\frac {1.5 \ L * 2.8 \ atm}{0.92 \ L}= P_2[/tex]
The units of liters cancel each other out.
[tex]\frac {1.5 * 2.8 \ atm}{0.92 }=P_2[/tex]
[tex]\frac {4.2}{0.92} \ atm= P_2[/tex]
[tex]4.565217391 \ atm = P_2[/tex]
The original measurements of pressure and volume have 2 significant figures, so our answer must have the same. For the number we calculated, that is the tenths place. The 6 in the hundredth place tells us to round the 5 up to a 6.
[tex]4.6 \ atm \approx P_2[/tex]
The pressure is approximately 4.6 atmospheres.
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