Answer to Question #97062 in General Chemistry for Brittany Wallace

Question

1. A sample of methane gas collected at a pressure of 226 mm Hg and a temperature of 286 K has a mass of 20.1 grams. The volume of the sample is __L.
2. A 24.3 gram sample of krypton gas has a volume of 927 milliliters at a pressure of 3.71 atm. The temperature of the Kr gas sample is ___°C.

Answer 1

Problem 1.

Methane has chemical formula $C{H_4}$ and molar mass ${M_{C{H_4}}} \approx 16.043[{{\rm{g}} \over {{\rm{mol}}}}]$ . To solve the problem we can use ideal gas law

$$pV = \nu RT$$

where $R \approx 8.314[{{\rm{J}} \over {{\rm{K}} \cdot {\rm{mol}}}}]$ - universal gas constant and $\nu = {m \over {{M_{C{H_4}}}}}$ - the amount of moles of methane.

Thus, the volume

$$V = {1 \over p}{m \over {{M_{C{H_4}}}}}RT$$

Let's do the calculations, using that $1[{\rm{mmHg}}] \approx 133.322[{\rm{Pa}}]$

$$V = {1 \over {226 \cdot 133.322[{\rm{Pa}}]}}{{20.1[{\rm{g}}]} \over {16.043[{{\rm{g}} \over {{\rm{mol}}}}]}}8.314[{{\rm{J}} \over {{\rm{K}} \cdot {\rm{mol}}}}] \cdot 286[{\rm{K}}] \approx 0.0989[{{\rm{m}}^3}] \approx 98.87[{\rm{L}}]$$

Problem 2.

Krypton $Kr$ has a molar mass ${M_{Kr}} \approx 83.798[{{\rm{g}} \over {{\rm{mol}}}}]$. Using the same ideal gas law we get for the temperature

$$T = {{pV} \over {\nu R}} = {{{M_{Kr}}} \over m}{{pV} \over R}$$

Let's do the calculations, using that $1[{\rm{atm}}] = 101325[{\rm{Pa}}]$

$$T = {{83.798[{{\rm{g}} \over {{\rm{mol}}}}]} \over {24.3[{\rm{g}}]}}{{3.71 \cdot 101325[{\rm{Pa}}] \cdot 927 \cdot {{10}^{ - 6}}[{{\rm{m}}^{\rm{3}}}]} \over {8.314[{{\rm{J}} \over {{\rm{K}} \cdot {\rm{mol}}}}]}} \approx 144.54[{\rm{K}}] \approx - 128.61[{\rm{^\circ C}}]$$