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Chemistry LibreTexts

Homework 48

  • Page ID
  • 10.41, 11.67

    10.41) What amount of energy needs to be removed from 0.75 pounds of silver to cool it from 110°F to 75°F? Assume silver has a density of 10.49 \[\frac{g}{cm^{3}}\]

    Use equation 10.5 on pg. 351 \[q= m\cdot C_{s}\cdot \Delta T\]

    From Table 10.2 on pg. 351, for silver Cs = 0.235 \[\frac{J}{g\cdot ^{\circ}C}\]

    First convert 0.75 pounds of silver to grams. \[0.75 lb\cdot 453.6\frac{g}{lb}=340.2 g\]

    Convert temperatures from °F to °C with the conversion °C = (X°F – 32°F)/1.8

    110°F – 32°F/1.8 = 43.3°C

    75°F – 32°F/1.8 = 23.9°C

    Using equation 10.5 \[q=340.2g\cdot 0.235\frac{J}{g\cdot C} \cdot \left ( 43.3 C-23.9 C \right )=1551 J\]

    11.67) Hydrogen gas produced by a chemical reaction is collected in the vapor space of a container holding water. The hydrogen partial pressure is 605 mmHg and the system temperature is 55 °C. What is the total pressure in the container? If 0.1 mole of H2 was produced in the reaction, what is the total volume of the gas above the water in the container in L? Use the gas constant R of 62.3637 \[\frac{L\cdot mmHg}{mol\cdot K}\]

    Answer, Part 1

    Knowing the system temperature is 55 °C, the vapor pressure of the water in the hydrogen/water vapor mixture can be determined. Using Table 11.3 on page 412, the water partial pressure is found to be 118.2 mmHg.

    From Equation 11.8 we know that PTotal = PH2O + PH2

    PH2 is given as 605 mmHg and PH2O was found to be 118.2 mmHg

    PTotal = 605 mmHg + 118.2 mmHg = 723.2 mmHg

    Answer, Part 2

    It is given there is 0.1 mole of H2 and hydrogen partial pressure 605 mmHg. In the first part of this problem, the total system pressure was found to be 723.2 mmHg.

    Solving Equation 11.10 for the total number of moles, we find \[n_{total}=\left ( \frac{n_{H_{2}}}{P_{H_{2}}} \right )\cdot P_{total}\]

    Therefore \[n_{total}=\left ( \frac{0.1 mol_{H_{2}}}{605mmHg_} \right )\cdot 723.2 mmHg_=0.12 mol\]

    From the Ideal Gas Law, Equation 11.5, we know PV=ntotalRT

    Solving for V, we find \[V=n_{total}\left ( \frac{RT}{P} \right )\]

    Convert the system temperature to Kelvin. K = 55 °C + 273.2 = 328.2 K

    Therefore V is:

    \[V=0.12mol\left ( \frac{62.3637\frac{L\cdot mmHg}{mol\cdot K}\cdot 328.2K}{723.2mmHg} \right )=3.4L\]

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