Question

a sample of a gas mixture contains the following quantities of three gases.

the sample has:
volume = 2.50 l
temperature = 16.6 °c

what is the partial pressure for each gas, in mmhg?
what is the total pressure in the flask?

co mmhg
co₂ mmhg
sf₆ mmhg
total mmhg

Explanation:

Step1: Calculate the number of moles for each gas

• For \(CO\):

The molar mass of \(CO\) is \(M_{CO}=(12.01 + 16.00)\space g/mol=28.01\space g/mol\).
Using the formula \(n=\frac{m}{M}\), where \(m = 1.61\space g\) and \(M = 28.01\space g/mol\), we have \(n_{CO}=\frac{1.61\space g}{28.01\space g/mol}\approx0.0575\space mol\).

• For \(CO_{2}\):

The molar mass of \(CO_{2}\) is \(M_{CO_{2}}=(12.01+2\times16.00)\space g/mol = 44.01\space g/mol\).
Using \(n=\frac{m}{M}\), with \(m = 2.41\space g\) and \(M = 44.01\space g/mol\), we get \(n_{CO_{2}}=\frac{2.41\space g}{44.01\space g/mol}\approx0.0548\space mol\).

• For \(SF_{6}\):

The molar mass of \(SF_{6}\) is \(M_{SF_{6}}=(32.07 + 6\times19.00)\space g/mol=146.07\space g/mol\).
Using \(n=\frac{m}{M}\), with \(m = 3.97\space g\) and \(M = 146.07\space g/mol\), we obtain \(n_{SF_{6}}=\frac{3.97\space g}{146.07\space g/mol}\approx0.0272\space mol\).

Step2: Convert temperature to Kelvin

The temperature \(T = 16.6^{\circ}C\). Using the formula \(T(K)=T(^{\circ}C)+273.15\), we have \(T=(16.6 + 273.15)\space K=289.75\space K\).

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

The ideal gas constant \(R = 62.36\space L\cdot mmHg/(mol\cdot K)\) and \(V = 2.50\space L\).

• For \(CO\):

From \(P=\frac{nRT}{V}\), substituting \(n = 0.0575\space mol\), \(R = 62.36\space L\cdot mmHg/(mol\cdot K)\), \(T = 289.75\space K\) and \(V = 2.50\space L\)
\(P_{CO}=\frac{0.0575\times62.36\times289.75}{2.50}\space mmHg\)
\(P_{CO}=\frac{0.0575\times62.36\times289.75}{2.50}\approx410\space mmHg\)

• For \(CO_{2}\):

Substituting \(n = 0.0548\space mol\), \(R = 62.36\space L\cdot mmHg/(mol\cdot K)\), \(T = 289.75\space K\) and \(V = 2.50\space L\)
\(P_{CO_{2}}=\frac{0.0548\times62.36\times289.75}{2.50}\space mmHg\)
\(P_{CO_{2}}=\frac{0.0548\times62.36\times289.75}{2.50}\approx390\space mmHg\)

• For \(SF_{6}\):

Substituting \(n = 0.0272\space mol\), \(R = 62.36\space L\cdot mmHg/(mol\cdot K)\), \(T = 289.75\space K\) and \(V = 2.50\space L\)
\(P_{SF_{6}}=\frac{0.0272\times62.36\times289.75}{2.50}\space mmHg\)
\(P_{SF_{6}}=\frac{0.0272\times62.36\times289.75}{2.50}\approx195\space mmHg\)

Step4: Calculate the total pressure

Using Dalton's law of partial pressures \(P_{total}=P_{CO}+P_{CO_{2}}+P_{SF_{6}}\)
\(P_{total}=(410 + 390+195)\space mmHg = 995\space mmHg\)

Answer:

CO: \(410\space mmHg\)
\(CO_{2}\): \(390\space mmHg\)
\(SF_{6}\): \(195\space mmHg\)
total: \(995\space mmHg\)