Question

question 5
a bubble released by a scuba diver at 10.0 m under water has a volume of 10.2 ml. at this depth
the pressure is 3.0 atm and the temperature is 15.0 ^ { circ } c. the bubble rises and reaches the surface
where the temperature is 25.0 ^ { circ } c and the pressure is 1.0 atm.
what is the volume of the bubble at the surface of the water?

question 6
a child is holding a balloon that contains 2.5 mol of helium gas and has a volume of 1.46 l. at the
surface of the ground the temperature is 32.0 ^ { circ } c and the pressure is 0.989 atm. the child lets the
balloon go and it rises into the atmosphere. the balloon gets to an altitude where the temperature
is - 5.0 ^ { circ } c. if the volume of the balloon is 4.25 l at this altitude, what is the pressure?

question 7
a 15.0 l container is filled with methane, ch _ { 4 }, gas at 50.0 ^ { circ } c and 2.5 atm. what is the mass of the
methane in the container?

Explanation:

Step1: Convert temperatures to Kelvin

For Question 5:

• Initial temperature \(T_1 = 15.0+273.15=288.15\space K\)
• Final temperature \(T_2 = 25.0 + 273.15=298.15\space K\)

For Question 6:

• Initial temperature \(T_1=32.0 + 273.15 = 305.15\space K\)
• Final temperature \(T_2=-5.0+273.15 = 268.15\space K\)

For Question 7:

• Temperature \(T=50.0+273.15 = 323.15\space K\)

Step2: Apply gas laws

Question 5:

Use the combined gas law \(\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\)
Given \(P_1 = 3.0\space atm\), \(V_1 = 10.2\space mL\), \(T_1=288.15\space K\), \(P_2 = 1.0\space atm\), \(T_2=298.15\space K\)
Rearrange for \(V_2\):
\(V_2=\frac{P_1V_1T_2}{P_2T_1}\)
\(V_2=\frac{3.0\times10.2\times298.15}{1.0\times288.15}\)
\(V_2=\frac{30.6\times298.15}{288.15}\)
\(V_2=\frac{9123.39}{288.15}\approx31.7\space mL\)

Question 6:

Use the combined gas law \(\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\)
Given \(n\) (moles of gas) is constant (since it's the same amount of gas in the balloon). Here \(P_1 = 0.989\space atm\), \(V_1 = 1.46\space L\), \(T_1 = 305.15\space K\), \(V_2=4.25\space L\), \(T_2 = 268.15\space K\)
Rearrange for \(P_2\):
\(P_2=\frac{P_1V_1T_2}{V_2T_1}\)
\(P_2=\frac{0.989\times1.46\times268.15}{4.25\times305.15}\)
First calculate numerator: \(0.989\times1.46\times268.15=(1.44494)\times268.15\approx387.5\)
Denominator: \(4.25\times305.15 = 1296.8875\)
\(P_2=\frac{387.5}{1296.8875}\approx0.30\space atm\)

Question 7:

Use the ideal gas law \(PV = nRT\), where \(R = 0.0821\space L\cdot atm/(mol\cdot K)\), \(P = 2.5\space atm\), \(V=15.0\space L\), \(T = 323.15\space K\)
First find \(n\) (moles of \(CH_4\)): \(n=\frac{PV}{RT}\)
\(n=\frac{2.5\times15.0}{0.0821\times323.15}\)
\(n=\frac{37.5}{26.53}\approx1.41\space mol\)
Molar mass of \(CH_4\) is \(M=(12.01 + 4\times1.008)=16.042\space g/mol\)
Mass \(m=n\times M\)
\(m = 1.41\times16.042\approx22.6\space g\)

Answer:

Question 5: \(31.7\space mL\)
Question 6: \(0.30\space atm\)
Question 7: \(22.6\space g\)