Chemistry Gas Law Questions – Answers + Explanations Bundle

Boyle’s Law states that \(P_1V_1 = P_2V_2\). Substituting the given values:
\(632.4 \times 5 = 356.8 \times V_2\). Solving for \(V_2\),
\(V_2 = \frac{632.4 \times 5}{356.8} \approx 8.86 \text{ L}\).

Boyle’s Law states that \(P_1V_1 = P_2V_2\). Substituting the given values:
\(P_1 \times 596 = 252.3 \times 323\). Solving for \(P_1\),
\(P_1 = \frac{252.3 \times 323}{596} \approx 136.63 \text{ atm}\).

Gay-Lussac’s Law states that \(\frac{P_1}{T_1} = \frac{P_2}{T_2}\). Substituting the given values:
\(\frac{86}{378} = \frac{P_2}{584}\). Solving for \(P_2\),
\(P_2 = \frac{86 \times 584}{378} \approx 132.87 \text{ atm}\).

Boyle’s Law states that \(P_1V_1 = P_2V_2\). Substituting the given values:
\(225 \times 22.5 = 45 \times V_2\). Solving for \(V_2\),
\(V_2 = \frac{225 \times 22.5}{45} = 112.5 \text{ L}\).

Charles’s Law states that \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\). Temperatures must be converted to Kelvin:
\(T_1 = 52 + 273 = 325 \text{ K}, T_2 = 58 + 273 = 331 \text{ K}\). Substituting:
\(\frac{V_1}{325} = \frac{795.6}{331}\). Solving for \(V_1\),
\(V_1 = \frac{325 \times 795.6}{331} \approx 781.42 \text{ L}\).

STP conditions are \(P_2 = 760 \text{ mmHg}\) and \(T_2 = 273 \text{ K}\). The Combined Gas Law is \(\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\). Convert temperature to Kelvin:
\(T_1 = 25.0 + 273 = 298 \text{ K}\). Substituting the given values:
\(\frac{600.0 \times 240.0}{298} = \frac{760 \times V_2}{273}\). Solving for \(V_2\):
\(V_2 = \frac{600.0 \times 240.0 \times 273}{298 \times 760} \approx 188.8 \text{ mL}\).

The Combined Gas Law is \(\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\). Convert temperatures to Kelvin:
\(T_1 = 60.0 + 273 = 333 \text{ K}\), \(T_2 = 27 + 273 = 300 \text{ K}\). Substituting the given values:
\(\frac{1.00 \times 4.20}{333} = \frac{P_2 \times 5.00}{300}\). Solving for \(P_2\):
\(P_2 = \frac{1.00 \times 4.20 \times 300}{333 \times 5.00} \approx 0.76 \text{ atm}\).

The Combined Gas Law is \(\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\). Convert pressures to the same unit:
\(P_2 = 1.00 \text{ atm} = 760 \text{ torr}\). Convert temperature to Kelvin:
\(T_1 = 77.0 + 273 = 350 \text{ K}\). Substituting the given values:
\(\frac{600.0 \times 500.0}{350} = \frac{760 \times 400.0}{T_2}\). Solving for \(T_2\):
\(T_2 = \frac{760 \times 400.0 \times 350}{600.0 \times 500.0} \approx 354.7 \text{ K}\), which is \(354.7 – 273 \approx 81.7^\circ\text{C}\).

The Combined Gas Law is \(\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\). Temperatures must be in Kelvin:
\(T_1 = 26.0 + 273 = 299.0 \text{ K}\). Substituting the given values:
\(\frac{660 \times 280.0}{299.0} = \frac{940 \times 440.0}{T_2}\). Solving for \(T_2\):
\(T_2 = \frac{940 \times 440.0 \times 299.0}{660 \times 280.0} \approx 693.6 \text{ K}\), which is \(693.6 – 273 \approx 420.6^\circ\text{C}\).

Charles’s Law states that \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\). Temperatures must be in Kelvin:
\(T_2 = 34.8 + 273 = 307.8 \text{ K}\). Substituting the given values:
\(\frac{598.7}{T_1} = \frac{399.1}{307.8}\). Solving for \(T_1\),
\(T_1 = \frac{598.7 \times 307.8}{399.1} \approx 461.8 \text{ K}\), which is \(461.8 – 273 \approx 188.8^\circ\text{C}\).

Charles’s Law states that \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\). Substituting the given values:
\(\frac{V_1}{478} = \frac{2.047}{346}\). Solving for \(V_1\),
\(V_1 = \frac{2.047 \times 478}{346} \approx 2.826 \text{ L}\).

Boyle’s Law states that \(P_1V_1 = P_2V_2\). Substituting the given values:
\(3.86 \times 747.2 = P_2 \times 465.8\). Solving for \(P_2\),
\(P_2 = \frac{3.86 \times 747.2}{465.8} \approx 6.19 \text{ atm}\).