A sample of air occupies 4.40 L when the pressure is 2.60 atm.

Chemistry

1 answer:

julia-pushkina [17]2 years ago

8 0

Explanation:

Given that,

Initial volume, V₁ = 4.40 L

Initial pressure, P₁ = 2.6 atm

(a) Final pressure, P₂ = 6.2 atm

As per the relation,

$P_1V_1=P_2V_2\\\\V_2=\dfrac{P_1V_1}{P_2}\\\\V_2=\dfrac{2.6\times 4.4}{6.2}\\\\V_2=1.84\ L$

(b) Again using the above relation,

$P_2=\dfrac{P_1V_1}{V_2}\\\\P_2=\dfrac{2.6\times 4.4}{0.0290 }\\\\P_2=394.48\ atm$

Hence, this is the required ssolution.

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A gas is initially confined to a 1.00 L vessel at 3.00 atm of pressure. A valve connecting the 1.00 L vessel to a 3.00 L vessel

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Answer:

c. 0.750 atm .

Explanation:

Hello!

In this case, since the two vessels have different volume, we can see that the gas is initially at 3.00 atm into the 1.00-L vessel, but next, it is allowed to move towards the 3.00-L vessel, meaning that the final volume wherein the gas is located, is 4.00 L; therefore, we use the Boyle's law to compute the final pressure:

$P_2V_2=P_1V_1\\\\P_2=\frac{P_1V_1}{V_2} \\\\P_2=\frac{3.00atm*1.00L}{4.00L}\\\\P_2=0.750atm$