Question

question 3
a 25.6 l tank of nitrogen gas is at 22.0 ^ { circ } mathrm { c } and 1.2 atm. if the temperature stays at 22.0 ^ { circ } mathrm { c } and the volume is compressed to 16.0 l, what is the new pressure?
question 4
a balloon containing helium gas has a volume of 1.43 l at room temperature 25 ^ { circ } mathrm { c }. the balloon is heated to a temperature of 83 ^ { circ } mathrm { c }. assuming no change in pressure, what is the new volume of the balloon?
question 5
a sample of gas is trapped in a rigid container at 1.2 atm and 31 ^ { circ } mathrm { c }. if the container is heated until the pressure in the container is 3.98 atm, what is the temperature of the gas in the container (in ^ { circ } mathrm { c } )?

Explanation:

Question 3

Step1: Identify the gas law

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

Step2: Rearrange for \(P_2\)

\(P_2=\frac{P_1V_1}{V_2}\)

Step3: Substitute values

\(P_1 = 1.2\) atm, \(V_1=25.6\) L, \(V_2 = 16.0\) L.
\(P_2=\frac{1.2\times25.6}{16.0}\)
\(P_2=\frac{30.72}{16.0}=1.92\) atm

Step1: Identify the gas law

Since pressure is constant, use Charles's Law \(\frac{V_1}{T_1}=\frac{V_2}{T_2}\)

Step2: Convert temperatures to Kelvin

\(T_1=25 + 273=298\) K, \(T_2=83+ 273 = 356\) K, \(V_1 = 1.43\) L

Step3: Rearrange for \(V_2\)

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

Step4: Substitute values

\(V_2=\frac{1.43\times356}{298}\)
\(V_2=\frac{509.08}{298}\approx1.71\) L

Step1: Identify the gas law

Since volume is constant (rigid container), use Gay - Lussac's Law \(\frac{P_1}{T_1}=\frac{P_2}{T_2}\)

Step2: Convert initial temperature to Kelvin

\(T_1=31 + 273=304\) K, \(P_1 = 1.2\) atm, \(P_2=3.98\) atm

Step3: Rearrange for \(T_2\)

\(T_2=\frac{P_2T_1}{P_1}\)

Step4: Substitute values

\(T_2=\frac{3.98\times304}{1.2}\)
\(T_2=\frac{1211.92}{1.2}\approx1009.93\) K

Step5: Convert back to Celsius

\(T_2=1009.93-273\approx737\) \(^{\circ}\text{C}\)

Answer:

\(1.92\)

Question 4