Question

a 3.00 l sample of a gas at a pressure of 4.00 atm is compressed to 2.00 l at a constant temperature. what is the pressure of the gas?
2.00 atm
6.00 atm
12.0 atm
24.0 atm
question 3
8 pts
a mole of gas at 0.0°c and 760. torr occupies 22.4 l. what is the volume of this gas at 25.°c and 760. torr?
24.5 l
0.0 l
20.5 l
19.8 l

Explanation:

Step1: Identify the gas law

Since temperature is constant, use Boyle's Law \(P_1V_1 = P_2V_2\).

Step2: Substitute the values

Given \(P_1 = 4.00\ atm\), \(V_1=3.00\ L\), \(V_2 = 2.00\ L\). Rearrange for \(P_2\): \(P_2=\frac{P_1V_1}{V_2}\).

Step3: Calculate \(P_2\)

\(P_2=\frac{4.00\times3.00}{2.00}=6.00\ atm\)

Step1: Identify the gas law

Since pressure is constant, use Charles's Law \(\frac{V_1}{T_1}=\frac{V_2}{T_2}\). Convert temperatures to Kelvin: \(T_1 = 0.0 + 273.15=273.15\ K\), \(T_2=25.0 + 273.15 = 298.15\ K\), \(V_1 = 22.4\ L\).

Step2: Rearrange for \(V_2\)

\(V_2=\frac{V_1T_2}{T_1}\)

Step3: Calculate \(V_2\)

\(V_2=\frac{22.4\times298.15}{273.15}\approx24.5\ L\)

Answer:

6.00 atm

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