Thermochemistry and Thermodynamics

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Physical Chemistry › Thermochemistry and Thermodynamics

Questions 1 - 10

A sample of an ideal gas is initially at a volume of $3.5 \textup{ L}$ . The gas expands to a volume of $7.0 \textup{ m}^3$ when $2.0 \textup{ J}$ of heat is applied to the system against a constant external pressure of $0.50 \textup{ Pa}$ . Calculate the change in internal energy for this gas.

$1.0\textup{ Pa}\cdot \textup{m}^{3}=1.0 \textup{ J}$

$0.25\textup{ J}$

$3.75\textup{ J}$

$2.25\textup{ J}$

$5.50\textup{ J}$

$1.20\textup{ J}$

Explanation

The expression for the relationship between heat () and work () is change in internal energy ( $\Delta U$ ):

$\Delta U=q+W$

The work () done by a system is:

$W=-p_{ex}\Delta V$

$\Delta V=V_{f}-V_{i}$ with $V_{f}=final\ volume$ and $V_{i}=initial\ volume$

Plugging work () into the internal energy equation gives:

$\Delta U=q-p_{ex}\Delta V$

Plugging the values given into the internal energy equation gives:

$\Delta U=2.0J-0.5Pa\times (7.0m^{3}-3.5m^{3})$

$\Delta U=2.0J-0.5Pa\times 3.5m^{3}$

$\Delta U=2.0J-1.75J=0.25J$

$1.0\textup{ Pa}\cdot \textup{m}^{3}=1.0 \textup{ J}$

$0.25\textup{ J}$

$3.75\textup{ J}$

$2.25\textup{ J}$

$5.50\textup{ J}$

$1.20\textup{ J}$

Explanation

The expression for the relationship between heat () and work () is change in internal energy ( $\Delta U$ ):

$\Delta U=q+W$

The work () done by a system is:

$W=-p_{ex}\Delta V$

$\Delta V=V_{f}-V_{i}$ with $V_{f}=final\ volume$ and $V_{i}=initial\ volume$

Plugging work () into the internal energy equation gives:

$\Delta U=q-p_{ex}\Delta V$

Plugging the values given into the internal energy equation gives:

$\Delta U=2.0J-0.5Pa\times (7.0m^{3}-3.5m^{3})$

$\Delta U=2.0J-0.5Pa\times 3.5m^{3}$

$\Delta U=2.0J-1.75J=0.25J$

$1.0\textup{ Pa}\cdot \textup{m}^{3}=1.0 \textup{ J}$

$0.25\textup{ J}$

$3.75\textup{ J}$

$2.25\textup{ J}$

$5.50\textup{ J}$

$1.20\textup{ J}$

Explanation

The expression for the relationship between heat () and work () is change in internal energy ( $\Delta U$ ):

$\Delta U=q+W$

The work () done by a system is:

$W=-p_{ex}\Delta V$

$\Delta V=V_{f}-V_{i}$ with $V_{f}=final\ volume$ and $V_{i}=initial\ volume$

Plugging work () into the internal energy equation gives:

$\Delta U=q-p_{ex}\Delta V$

Plugging the values given into the internal energy equation gives:

$\Delta U=2.0J-0.5Pa\times (7.0m^{3}-3.5m^{3})$

$\Delta U=2.0J-0.5Pa\times 3.5m^{3}$

$\Delta U=2.0J-1.75J=0.25J$

In an reaction, the products are more stable than the reactants; in an reaction the reactants are more stable than the products.

exergonic . . . endergonic

endergonic . . . exergonic

endergonic . . . endergonic

exergonic . . . exergonic

Explanation

Exergonic reactions release energy; therefore, the energy of products is lower than that of the reactants. Endergonic reactions consume energy; therefore, the energy of products is greater than that of the reactants. In other words, exergonic reactions are spontaneous, while endergonic reactions are nonspontaneous, and require the net input of energy to drive the reaction.

In an reaction, the products are more stable than the reactants; in an reaction the reactants are more stable than the products.

exergonic . . . endergonic

endergonic . . . exergonic

endergonic . . . endergonic

exergonic . . . exergonic

Explanation

In an reaction, the products are more stable than the reactants; in an reaction the reactants are more stable than the products.

exergonic . . . endergonic

endergonic . . . exergonic

endergonic . . . endergonic

exergonic . . . exergonic

Explanation

In an exergonic reaction, products will have Gibbs free energy and the reaction is .

lower . . . spontaneous

lower . . . nonspontaneous

higher . . . spontaneous

higher . . . nonspontaneous

Explanation

Exergonic reaction suggests that the Gibbs free energy is negative. Since the change in Gibbs free energy is defined as Gibbs free energy of products - Gibbs free energy of reactants, a negative change in Gibbs free energy suggests that the products have a lower Gibbs free energy than reactants. A reaction is spontaneous if it has negative Gibbs free energy; therefore, exergonic reactions are always spontaneous. This is because the reaction is producing a more stable product (lower energy) from a less stable reactant (higher energy).

At $270 \textup{ K}$ , a $0.1446 \textup{ mol}$ sample of argon gas expands reversibly from a confined space of $2.1 \textup{ m}^3$ to $3.4 \textup{ m}^3$ . Calculate the work done.

$156 \textup{ J}$

$312\textup{ J}$

$78\textup{ J}$

$300\textup{ J}$

$1.533\textup{ J}$

Explanation

The process as described by the question is an isothermal process. An isothermal process has a constant temperature, therefore, there is no change in temperature. For an isothermal reversible process, the work done by the system is:

$W=-nRTln(\frac{v_{f}}{v_{i}})$

Plugging the values given into the equation gives:

$W=-(0.1446\ mol)\times (8.3145\ \frac{J}{K\cdot mol})\times (270\ K)\times (ln\frac{3.4m^{3}}{2.1m^{3}})=156\ J$

At $270 \textup{ K}$ , a $0.1446 \textup{ mol}$ sample of argon gas expands reversibly from a confined space of $2.1 \textup{ m}^3$ to $3.4 \textup{ m}^3$ . Calculate the work done.

$156 \textup{ J}$

$312\textup{ J}$

$78\textup{ J}$

$300\textup{ J}$

$1.533\textup{ J}$