Question

module 29 | extra practice: gas laws
question 1
a 0.85 mol sample of a gas at 27°c is in a rigid container with a volume of 1.65 l. what is the pressure of the gas in atm?

question 2
a sample of gas occupies 4.50 l at 45°c and a pressure of 1.25 atm. if the temperature is increased to 65°c and the volume is held constant, what is the new pressure?
1.80 atm
1.33 atm
1.18 atm
0.865 atm
5.98 atm

Explanation:

Step1: Convert temperature to Kelvin

The formula to convert Celsius to Kelvin is \(T(K)=T(^{\circ}C)+273.15\).
For the first problem, \(T = 27+273.15=300.15\space K\).
For the second problem, \(T_1 = 45 + 273.15=318.15\space K\) and \(T_2=65 + 273.15 = 338.15\space K\).

Step2: Use the ideal gas law for problem 1

The ideal gas law is \(PV=nRT\), where \(P\) is pressure, \(V\) is volume, \(n\) is the number of moles, \(R\) is the ideal gas constant (\(R = 0.0821\space L\cdot atm/(mol\cdot K)\)), and \(T\) is temperature.
We need to solve for \(P\), so \(P=\frac{nRT}{V}\).
Substitute \(n = 0.85\space mol\), \(R=0.0821\space L\cdot atm/(mol\cdot K)\), \(T = 300.15\space K\), and \(V=1.65\space L\) into the formula:
\(P=\frac{0.85\times0.0821\times300.15}{1.65}\)
\(P=\frac{0.85\times24.642315}{1.65}\)
\(P=\frac{20.94596775}{1.65}\approx12.7\space atm\)

Step3: Use Gay - Lussac's law for problem 2

Gay - Lussac's law is \(\frac{P_1}{T_1}=\frac{P_2}{T_2}\) (since \(V\) is constant).
We know \(P_1 = 1.25\space atm\), \(T_1 = 318.15\space K\), and \(T_2=338.15\space K\).
Solve for \(P_2\): \(P_2=\frac{P_1T_2}{T_1}\)
\(P_2=\frac{1.25\times338.15}{318.15}\)
\(P_2=\frac{422.6875}{318.15}\approx1.33\space atm\)

Answer:

For question 1: \(12.7\space atm\)
For question 2: B. \(1.33\space atm\)