,DOWNLOAD THE Test Bank for Elements of Physical Chemistry 7th Edition
f f f f f f f f f f
Atkins
Atkins & de Paula: Elements of Physical Chemistry 7e f f f f f f f f
Test bank: Focus 02 f f f
Type: multiple choice question Title:
f f f f
Focus 02 - Question 01
f f f f f
01) Calculate the expansion work done on the system when exactly 1 mol of solid ammonium f f f f f f f f f f f f f f
chloride, NH4Cl, decomposes completely to yield gaseous ammonia, NH3 and hydrogen
f f f f f f f f f f f
chloride, HCl at a temperature of 1250 K. Treat the expansion as irreversible and the gases
f f f f f f f f f f f f f f f f
formed as perfect.
f f f
Feedback: The work done on a system when it expands against a constant pressure is given by f f f f f f f f f f f f f f f f
eqn 2A.1a
f f
𝑤 = −𝑝exΔ𝑉 f f
The decomposition of exactly 1 mol of solid ammonium chloride
f f f f f f f f f
NH4Cl(s) NH3(g) + HCl(g) yields 2 mol of gas. We may assume that the volume of the solid
f f f f f f f f f f f f f f f f f f
ammonium chloride is negligible in comparison with the volume of the gas produced. The
f f f f f f f f f f f f f f
change in volume on decomposition is therefore
f f f f f f f
Δ𝑉 = 𝑉f − 𝑉i ≈ 𝑉f f f
f
f
f
f
Treating the gas as perfect, so that f f f f f f
𝑛𝑅𝑇
𝑉f
𝑝ex
=
then f
𝑛𝑅𝑇
𝑤 = −𝑝ex × f f
𝑝ex f
= −𝑛𝑅𝑇 f
= −(2.00 mol) × (8.3145 J K−1mol−1) × (1250 K) f f f f f f f f f
= −20.8 × 103J f f f
= −𝟐𝟎. 𝟖 𝐤𝐉 f f f
Page reference: 41 f f
a. –15.4 kJ f
b. –4.96 kJ f
c. –16.6 kJ f
*d. –20.8 kJf f
Type: multiple choice question Title:
f f f f
Focus 02 - Question 02
f f f f f
02) Calculate the heat transferred to the system when 1.00 mol of a perfect gas expands f f f f f f f f f f f f f f
reversibly at a constant temperature of 25°C so that its volume doubles.
f f f f f f f f f f f f
Feedback: For the isothermal expansion of a perfect gas, eqn 2B.4 shows that the heat transferred
f f f f f f f f f f f f f f f
and work done are related by
f f f f f f
𝑞 = −𝑤 f f
and so, using eqn 2A.2 for a reversible expansion, we obtain eqn 2B.5,
f f f f f f f f f f f f
𝑞 = −𝑛𝑅𝑇𝑙𝑛𝑉f/𝑉i f f
For this expansion,
f f
𝑉f/𝑉i = 2 f
f
so that f
𝑞 = −(1.00 mol) × (8.3145 J K−1mol−1) × (273 + 25)K × ln 2 f f f f f f f f f f f f f f
= −1.72 × 103J f f f
= −𝟏. 𝟕𝟐 𝐤𝐉 𝐦𝐨𝐥−𝟏 f f f f
Page reference: 43, 51 f f f
a. –144 J f
b. –746 J f
*c. –1.72 kJ mol–1
f f f
d. 2.48 kJ
f f
Type: multiple choice question Title:
f f f f
Focus 02 - Question 03
f f f f f
08) The constant pressure molar heat capacity of zinc is 25.40 J K–1 mol–1 at 298 K. Calculate
f f f f f f f f f f f f
f f
f f f
the constant pressure specific heat capacity of zinc at this temperature.
f f f f f f f f f f f
© Oxford University Press, 2017.
f f f f
mynursytest.store
,DOWNLOAD THE Test Bank for Elements of Physical Chemistry 7th Edition
f f f f f f f f f f
Atkins
Atkins & de Paula: Elements of Physical Chemistry 7e f f f f f f f f
Feedback: Specific heat capacity is defined as the heat capacity per unit mass of sample f f f f f f f f f f f f f f
𝐶s = 𝐶/𝑚 f
f
and so, f
𝐶s = 𝐶𝑝,m/𝑀 f
f
For zinc, the molar mass is
f f f f f
𝑀 = 65.37 g mol−1 f f f f
so that
f f
𝐶s = 𝐶𝑝,m/𝑀 f
f
= (25.40 J K−1mol−1) / (65.37 × 10−3kg mol−1) f f f f f f f f
= 𝟑𝟖𝟖. 𝟔 𝐉 𝐊−𝟏𝐤𝐠−𝟏 f f f f
Page reference: 48 f f
a. 1.660 kJ K–1 kg–1 f f
f
*b. 388.6 J K–1 kg–1
f f f
f
c. 17.09 J K–1 kg–1 f f
f
d. 33.71 J K–1 kg–1 f f
f
Type: multiple choice question Title: f f f f
Focus 02 - Question 04
f f f f f
03) The molar heat capacity of solid aluminium is 24.4 J K–1 mol–1 at 25°C. Calculate the
f f f f f f f f f f
f f
f f f
change in internal energy when 1.00 mol of solid aluminium is heated from a temperature of
f f f f f f f f f f f f f f f f
20°C to 30°C.
f f f
Feedback: Heat capacity is defined by eqn 2C.4a and so we can write f f f f f f f f f f f f
∆𝑈 = 𝐶𝑉Δ𝑇 f f
For a solid, f f
𝐶𝑝 ≈ 𝐶𝑉 f
f
and so we do not need to worry about the distinction between the heat capacity at constant volume
f f f f f f f f f f f f f f f f f
and pressure. We may also reasonably assume that the heat capacity does not vary over the
f f f f f f f f f f f f f f f f
range of temperature. A change in temperature from 20 °C to 30 °C means
f f f f f f f f f f f f f f
Δ𝑇 = +10 K f f f
Thus
Δ𝑈 = 𝑛𝐶mΔ𝑇 f f
= (1.00 mol) × (24.4 J K–1mol−1) × (10 K)
f f f f f f f f f
= 𝟐𝟒𝟒 𝐉 f f
Page reference: 54 f f
*a. 244 J f f
b. 24.4 J f
c. 171 J f
d. 327 J f
Type: multiple choice question Title: f f f f
Focus 02 - Question 05
f f f f f
04) The constant pressure molar heat capacity of ammonia, NH3, has been found to vary with
f f f f f f f f f f f f f f
temperature according to the relation
f f f f f
𝐶𝑝,m / J K−1mol−1 = 𝑎 + 𝑏𝑇 + 𝑐/𝑇2 f
f f f f f f f f
with a = 29.73, b = 25.1 10 K and c = –1.55 105 K2. Calculate the value of the constant
f
–3 –1 f f f f f f f
f f
f f f f f
f
f f f f f f
pressure molar heat capacity at 25.0 °C.
f f f f f f f
Feedback: The constant pressure molar heat capacity may be calculated by direct f f f f f f f f f f f
substitution. We must, however, remember to convert the temperature to units of kelvin. Thus
f f f f f f f f f f f f f f
𝑇 = (273.15 + 25.0)K = 298.15 K f f f f f f f
so that f
𝑐
𝐶𝑝,m /( J K mol ) = 𝑎 + 𝑏𝑇 + 2
−1 −1 f f f f f f f f
𝑇
= 29.73 + {(25.1 × 10−3 K−1) × (298.15 K)} − {1.55 × 105 K2/(298.15 K)2} f f f f f f f f f f f f f f f
= 35.47 f
Thus
© Oxford University Press, 2017.
f f f f
mynursytest.store
, DOWNLOAD THE Test Bank for Elements of Physical Chemistry 7th Edition
f f f f f f f f f f
Atkins
Atkins & de Paula: Elements of Physical Chemistry 7e f f f f f f f f
𝐶𝑝,m = 𝟑𝟓. 𝟒𝟕 𝐉 𝐊−𝟏𝐦𝐨𝐥−𝟏 f
f f f f
Page reference: 59 f f
a. 217.6 J K–1 mol–1 f f
f
b. 29.73 J K–1 mol–1 f f
f
*c. 35.47 J K–1 mol–1
f f f
f
d. 657.2 J K–1 mol–1
f f f
f
Type: multiple choice question Title: f f f f
Focus 02 - Question 06
f f f f f
05) In the calibration step of a thermochemistry experiment, a current of 117 mA, from a
f f f f f f f f f f f f f f
24.0 V source was allowed to flow through the electrical heater for 247 s and was found to result in
f f f f f f f f f f f f f f f f f f f
an increase in the temperature of the calorimeter and its contents of +1.25 K. Calculate the heat
f f f f f f f f f f f f f f f f f
capacity of the calorimeter and its contents.
f f f f f f f
Feedback: Heat capacity is defined by eqn 2B.1 f f f f f f f
𝐶 = 𝑞/Δ𝑇 f f
For an electrical heater, the amount of heat transferred is given by eqn 2B.3
f f f f f f f f f f f f f
𝑞cal = 𝐼𝒱𝑡 f
f
so that, if we combine these two expressions
f f f f f f f
𝐶 = 𝐼𝒱𝑡/ Δ𝑇 f f f
= (117 × 10−3A) × (24.0 V) × (247 s) / (1.25 K) f f f f f f f f f f f f
= 𝟓𝟓𝟓 𝐉 𝐊−𝟏 f f f
Page reference: 50 f f
a. 694 J K–1 f f
b. 277 J K–1 f f
*c. 555 J K–1 f f f
d. 867 J K–1
f f f
Type: multiple choice question Title: f f f f
Focus 02 - Question 07
f f f f f
06) The constant pressure molar heat capacity of methane, CH4, is 35.31 J K–1 mol–1 at
f f f f f f f f f f f f f f
temperatures close to 298 K. Calculate the enthalpy change when 2.00 mol of methane is
f f f f f f f f f f f f f f f
heated from a temperature of 278 K to 318 K.
f f f f f f f f f f
Feedback: We may calculate the change in enthalpy on heating by rearranging eqn 2B.5 f f f f f f f f f f f f f
Δ𝐻 = 𝐶Δ𝑇 f f
if we know that the heat capacity is constant and does not vary with temperature. We know the
f f f f f f f f f f f f f f f f f
constant pressure molar heat capacity, so that
f f f f f f f
𝐶 = 𝑛 𝐶𝑝,m f f f
Δ𝐻 = 𝑛 𝐶𝑝,m Δ𝑇 f f f
f
= (2.00 mol) × (35.31 J K−1mol−1) × (318 − 278) K f f f f f f f f f f f
= 2830 J = 𝟐. 𝟖𝟑 𝐤𝐉 f f f f f f
Page reference: 59 f f
a. 21.2 kJ f
b. 353 J f
c. 1.41 kJ f
*d. 2.83 kJ f f
Type: multiple choice question Title: f f f f
Focus 02 - Question 08
f f f f f
07) The constant pressure molar heat capacity of argon is 20.79 J K–1 mol–1 at 298 K. Predict the
f f f f f f f f f f f
f f
f f f f
value of the constant volume molar heat capacity of argon at this temperature.
f f f f f f f f f f f f f
Feedback: For a perfect gas, eqn 2D.7, shows that f f f f f f f f
𝐶𝑝,m − 𝐶𝑉,m = 𝑅 f
f
f
f
and hence f
𝐶𝑉,m = 𝐶𝑝,m − 𝑅 f
f
f
f
= (20.79 J K−1mol−1) − (8.3145 J K−1mol−1) f f f f f f f
= 𝟏𝟐. 𝟒𝟖 𝐉 𝐊−𝟏𝐦𝐨𝐥−𝟏 f f f f
© Oxford University Press, 2017.
f f f f
mynursytest.store
f f f f f f f f f f
Atkins
Atkins & de Paula: Elements of Physical Chemistry 7e f f f f f f f f
Test bank: Focus 02 f f f
Type: multiple choice question Title:
f f f f
Focus 02 - Question 01
f f f f f
01) Calculate the expansion work done on the system when exactly 1 mol of solid ammonium f f f f f f f f f f f f f f
chloride, NH4Cl, decomposes completely to yield gaseous ammonia, NH3 and hydrogen
f f f f f f f f f f f
chloride, HCl at a temperature of 1250 K. Treat the expansion as irreversible and the gases
f f f f f f f f f f f f f f f f
formed as perfect.
f f f
Feedback: The work done on a system when it expands against a constant pressure is given by f f f f f f f f f f f f f f f f
eqn 2A.1a
f f
𝑤 = −𝑝exΔ𝑉 f f
The decomposition of exactly 1 mol of solid ammonium chloride
f f f f f f f f f
NH4Cl(s) NH3(g) + HCl(g) yields 2 mol of gas. We may assume that the volume of the solid
f f f f f f f f f f f f f f f f f f
ammonium chloride is negligible in comparison with the volume of the gas produced. The
f f f f f f f f f f f f f f
change in volume on decomposition is therefore
f f f f f f f
Δ𝑉 = 𝑉f − 𝑉i ≈ 𝑉f f f
f
f
f
f
Treating the gas as perfect, so that f f f f f f
𝑛𝑅𝑇
𝑉f
𝑝ex
=
then f
𝑛𝑅𝑇
𝑤 = −𝑝ex × f f
𝑝ex f
= −𝑛𝑅𝑇 f
= −(2.00 mol) × (8.3145 J K−1mol−1) × (1250 K) f f f f f f f f f
= −20.8 × 103J f f f
= −𝟐𝟎. 𝟖 𝐤𝐉 f f f
Page reference: 41 f f
a. –15.4 kJ f
b. –4.96 kJ f
c. –16.6 kJ f
*d. –20.8 kJf f
Type: multiple choice question Title:
f f f f
Focus 02 - Question 02
f f f f f
02) Calculate the heat transferred to the system when 1.00 mol of a perfect gas expands f f f f f f f f f f f f f f
reversibly at a constant temperature of 25°C so that its volume doubles.
f f f f f f f f f f f f
Feedback: For the isothermal expansion of a perfect gas, eqn 2B.4 shows that the heat transferred
f f f f f f f f f f f f f f f
and work done are related by
f f f f f f
𝑞 = −𝑤 f f
and so, using eqn 2A.2 for a reversible expansion, we obtain eqn 2B.5,
f f f f f f f f f f f f
𝑞 = −𝑛𝑅𝑇𝑙𝑛𝑉f/𝑉i f f
For this expansion,
f f
𝑉f/𝑉i = 2 f
f
so that f
𝑞 = −(1.00 mol) × (8.3145 J K−1mol−1) × (273 + 25)K × ln 2 f f f f f f f f f f f f f f
= −1.72 × 103J f f f
= −𝟏. 𝟕𝟐 𝐤𝐉 𝐦𝐨𝐥−𝟏 f f f f
Page reference: 43, 51 f f f
a. –144 J f
b. –746 J f
*c. –1.72 kJ mol–1
f f f
d. 2.48 kJ
f f
Type: multiple choice question Title:
f f f f
Focus 02 - Question 03
f f f f f
08) The constant pressure molar heat capacity of zinc is 25.40 J K–1 mol–1 at 298 K. Calculate
f f f f f f f f f f f f
f f
f f f
the constant pressure specific heat capacity of zinc at this temperature.
f f f f f f f f f f f
© Oxford University Press, 2017.
f f f f
mynursytest.store
,DOWNLOAD THE Test Bank for Elements of Physical Chemistry 7th Edition
f f f f f f f f f f
Atkins
Atkins & de Paula: Elements of Physical Chemistry 7e f f f f f f f f
Feedback: Specific heat capacity is defined as the heat capacity per unit mass of sample f f f f f f f f f f f f f f
𝐶s = 𝐶/𝑚 f
f
and so, f
𝐶s = 𝐶𝑝,m/𝑀 f
f
For zinc, the molar mass is
f f f f f
𝑀 = 65.37 g mol−1 f f f f
so that
f f
𝐶s = 𝐶𝑝,m/𝑀 f
f
= (25.40 J K−1mol−1) / (65.37 × 10−3kg mol−1) f f f f f f f f
= 𝟑𝟖𝟖. 𝟔 𝐉 𝐊−𝟏𝐤𝐠−𝟏 f f f f
Page reference: 48 f f
a. 1.660 kJ K–1 kg–1 f f
f
*b. 388.6 J K–1 kg–1
f f f
f
c. 17.09 J K–1 kg–1 f f
f
d. 33.71 J K–1 kg–1 f f
f
Type: multiple choice question Title: f f f f
Focus 02 - Question 04
f f f f f
03) The molar heat capacity of solid aluminium is 24.4 J K–1 mol–1 at 25°C. Calculate the
f f f f f f f f f f
f f
f f f
change in internal energy when 1.00 mol of solid aluminium is heated from a temperature of
f f f f f f f f f f f f f f f f
20°C to 30°C.
f f f
Feedback: Heat capacity is defined by eqn 2C.4a and so we can write f f f f f f f f f f f f
∆𝑈 = 𝐶𝑉Δ𝑇 f f
For a solid, f f
𝐶𝑝 ≈ 𝐶𝑉 f
f
and so we do not need to worry about the distinction between the heat capacity at constant volume
f f f f f f f f f f f f f f f f f
and pressure. We may also reasonably assume that the heat capacity does not vary over the
f f f f f f f f f f f f f f f f
range of temperature. A change in temperature from 20 °C to 30 °C means
f f f f f f f f f f f f f f
Δ𝑇 = +10 K f f f
Thus
Δ𝑈 = 𝑛𝐶mΔ𝑇 f f
= (1.00 mol) × (24.4 J K–1mol−1) × (10 K)
f f f f f f f f f
= 𝟐𝟒𝟒 𝐉 f f
Page reference: 54 f f
*a. 244 J f f
b. 24.4 J f
c. 171 J f
d. 327 J f
Type: multiple choice question Title: f f f f
Focus 02 - Question 05
f f f f f
04) The constant pressure molar heat capacity of ammonia, NH3, has been found to vary with
f f f f f f f f f f f f f f
temperature according to the relation
f f f f f
𝐶𝑝,m / J K−1mol−1 = 𝑎 + 𝑏𝑇 + 𝑐/𝑇2 f
f f f f f f f f
with a = 29.73, b = 25.1 10 K and c = –1.55 105 K2. Calculate the value of the constant
f
–3 –1 f f f f f f f
f f
f f f f f
f
f f f f f f
pressure molar heat capacity at 25.0 °C.
f f f f f f f
Feedback: The constant pressure molar heat capacity may be calculated by direct f f f f f f f f f f f
substitution. We must, however, remember to convert the temperature to units of kelvin. Thus
f f f f f f f f f f f f f f
𝑇 = (273.15 + 25.0)K = 298.15 K f f f f f f f
so that f
𝑐
𝐶𝑝,m /( J K mol ) = 𝑎 + 𝑏𝑇 + 2
−1 −1 f f f f f f f f
𝑇
= 29.73 + {(25.1 × 10−3 K−1) × (298.15 K)} − {1.55 × 105 K2/(298.15 K)2} f f f f f f f f f f f f f f f
= 35.47 f
Thus
© Oxford University Press, 2017.
f f f f
mynursytest.store
, DOWNLOAD THE Test Bank for Elements of Physical Chemistry 7th Edition
f f f f f f f f f f
Atkins
Atkins & de Paula: Elements of Physical Chemistry 7e f f f f f f f f
𝐶𝑝,m = 𝟑𝟓. 𝟒𝟕 𝐉 𝐊−𝟏𝐦𝐨𝐥−𝟏 f
f f f f
Page reference: 59 f f
a. 217.6 J K–1 mol–1 f f
f
b. 29.73 J K–1 mol–1 f f
f
*c. 35.47 J K–1 mol–1
f f f
f
d. 657.2 J K–1 mol–1
f f f
f
Type: multiple choice question Title: f f f f
Focus 02 - Question 06
f f f f f
05) In the calibration step of a thermochemistry experiment, a current of 117 mA, from a
f f f f f f f f f f f f f f
24.0 V source was allowed to flow through the electrical heater for 247 s and was found to result in
f f f f f f f f f f f f f f f f f f f
an increase in the temperature of the calorimeter and its contents of +1.25 K. Calculate the heat
f f f f f f f f f f f f f f f f f
capacity of the calorimeter and its contents.
f f f f f f f
Feedback: Heat capacity is defined by eqn 2B.1 f f f f f f f
𝐶 = 𝑞/Δ𝑇 f f
For an electrical heater, the amount of heat transferred is given by eqn 2B.3
f f f f f f f f f f f f f
𝑞cal = 𝐼𝒱𝑡 f
f
so that, if we combine these two expressions
f f f f f f f
𝐶 = 𝐼𝒱𝑡/ Δ𝑇 f f f
= (117 × 10−3A) × (24.0 V) × (247 s) / (1.25 K) f f f f f f f f f f f f
= 𝟓𝟓𝟓 𝐉 𝐊−𝟏 f f f
Page reference: 50 f f
a. 694 J K–1 f f
b. 277 J K–1 f f
*c. 555 J K–1 f f f
d. 867 J K–1
f f f
Type: multiple choice question Title: f f f f
Focus 02 - Question 07
f f f f f
06) The constant pressure molar heat capacity of methane, CH4, is 35.31 J K–1 mol–1 at
f f f f f f f f f f f f f f
temperatures close to 298 K. Calculate the enthalpy change when 2.00 mol of methane is
f f f f f f f f f f f f f f f
heated from a temperature of 278 K to 318 K.
f f f f f f f f f f
Feedback: We may calculate the change in enthalpy on heating by rearranging eqn 2B.5 f f f f f f f f f f f f f
Δ𝐻 = 𝐶Δ𝑇 f f
if we know that the heat capacity is constant and does not vary with temperature. We know the
f f f f f f f f f f f f f f f f f
constant pressure molar heat capacity, so that
f f f f f f f
𝐶 = 𝑛 𝐶𝑝,m f f f
Δ𝐻 = 𝑛 𝐶𝑝,m Δ𝑇 f f f
f
= (2.00 mol) × (35.31 J K−1mol−1) × (318 − 278) K f f f f f f f f f f f
= 2830 J = 𝟐. 𝟖𝟑 𝐤𝐉 f f f f f f
Page reference: 59 f f
a. 21.2 kJ f
b. 353 J f
c. 1.41 kJ f
*d. 2.83 kJ f f
Type: multiple choice question Title: f f f f
Focus 02 - Question 08
f f f f f
07) The constant pressure molar heat capacity of argon is 20.79 J K–1 mol–1 at 298 K. Predict the
f f f f f f f f f f f
f f
f f f f
value of the constant volume molar heat capacity of argon at this temperature.
f f f f f f f f f f f f f
Feedback: For a perfect gas, eqn 2D.7, shows that f f f f f f f f
𝐶𝑝,m − 𝐶𝑉,m = 𝑅 f
f
f
f
and hence f
𝐶𝑉,m = 𝐶𝑝,m − 𝑅 f
f
f
f
= (20.79 J K−1mol−1) − (8.3145 J K−1mol−1) f f f f f f f
= 𝟏𝟐. 𝟒𝟖 𝐉 𝐊−𝟏𝐦𝐨𝐥−𝟏 f f f f
© Oxford University Press, 2017.
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