Question

a 5.73 - l flask at 25°c contains 0.0388 mol of n₂, 0.147 mol of co, and 0.0803 mol of h₂. what is the total pressure in the flask in atmospheres?
first calculate the pressure of each gas.
p_{n₂}= atm
p_{co}= atm
p_{h₂}= atm
now calculate the total pressure of the mixture of gases.
p_{total}= atm
what is the pressure of a mixture of 0.200 g of h₂, 1.00 g of n₂, and 0.820 g of ar in a container with a volume of 2.00 l at 20°c?
first calculate the partial pressure of each of the gases.
p_{h₂}= atm
p_{n₂}= atm
p_{ar}= atm

Explanation:

Step1: Convert temperature to Kelvin

For the first problem: \(T = 25+273 = 298\space K\)
For the second problem: \(T = 20 + 273=293\space K\)

Step2: Calculate the number of moles for the second problem

Molar mass of \(H_2\) is \(M_{H_2}=2\space g/mol\), \(n_{H_2}=\frac{0.200\space g}{2\space g/mol}=0.1\space mol\)
Molar mass of \(N_2\) is \(M_{N_2}=28\space g/mol\), \(n_{N_2}=\frac{1.00\space g}{28\space g/mol}\approx0.0357\space mol\)
Molar mass of \(Ar\) is \(M_{Ar}=40\space g/mol\), \(n_{Ar}=\frac{0.820\space g}{40\space g/mol}=0.0205\space mol\)

Step3: Use the ideal gas law \(PV = nRT\) to find partial pressures

For the first problem:

• \(P_{N_2}=\frac{n_{N_2}RT}{V}\), where \(R = 0.0821\space L\cdot atm/(mol\cdot K)\), \(n_{N_2}=0.0388\space mol\), \(V = 5.73\space L\), \(T = 298\space K\)

\(P_{N_2}=\frac{0.0388\times0.0821\times298}{5.73}\approx0.166\space atm\)

• \(P_{CO}=\frac{n_{CO}RT}{V}\), \(n_{CO}=0.147\space mol\)

\(P_{CO}=\frac{0.147\times0.0821\times298}{5.73}\approx0.631\space atm\)

• \(P_{H_2}=\frac{n_{H_2}RT}{V}\), \(n_{H_2}=0.0803\space mol\)

\(P_{H_2}=\frac{0.0803\times0.0821\times298}{5.73}\approx0.345\space atm\)

• \(P_{total}=P_{N_2}+P_{CO}+P_{H_2}=0.166 + 0.631+0.345 = 1.142\space atm\)

For the second problem:

• \(P_{H_2}=\frac{n_{H_2}RT}{V}\), \(n_{H_2}=0.1\space mol\), \(V = 2.00\space L\), \(T = 293\space K\)

\(P_{H_2}=\frac{0.1\times0.0821\times293}{2.00}\approx1.20\space atm\)

• \(P_{N_2}=\frac{n_{N_2}RT}{V}\), \(n_{N_2}\approx0.0357\space mol\)

\(P_{N_2}=\frac{0.0357\times0.0821\times293}{2.00}\approx0.429\space atm\)

• \(P_{Ar}=\frac{n_{Ar}RT}{V}\), \(n_{Ar}=0.0205\space mol\)

\(P_{Ar}=\frac{0.0205\times0.0821\times293}{2.00}\approx0.247\space atm\)

Answer:

For the first problem:
\(P_{N_2}\approx0.166\space atm\)
\(P_{CO}\approx0.631\space atm\)
\(P_{H_2}\approx0.345\space atm\)
\(P_{total}\approx1.14\space atm\)

For the second problem:
\(P_{H_2}\approx1.20\space atm\)
\(P_{N_2}\approx0.429\space atm\)
\(P_{Ar}\approx0.247\space atm\)