Test Review U4P1

If 1.000 cal = 4.184 Joules, how many Joules are in 41.84 cal?

• 75 J

• 175 J

• 100 J

• 1000 J

Answer explanation

Hot coffee in a vacuum flask (thermos) is an example of what type of system?

• isolated

• closed

• open

• universal

Answer explanation

An isolated system cannot exchange energy nor matter with the surroundings.

Which of the following are state functions?

1.Pressure (P)

2. Work (w)

3. Entropy (S)

4. Temperature (T)

• 3 only

• 1, 2, 3

• 1 only

• 1, 3, 4

Answer explanation

State function is a property whose value does not depend on the path taken to reach that specific value. Examples of state functions include mass, volume, enthalpy, entropy, internal energy, free energy, pressure, and temperature. In contrast, the two most common path functions are heat and work. 

In the formula ΔU = ΔE = q + w, work done by the system during expansion is ________.

• negative

• dependent on the system

• positive

• zero

Answer explanation

The sign conventions for path functions (like work and heat) are such that a negative sign is attributed to the system if it loses energy in the form of heat or does work.

Two metals of equal mass but with differing heat capacities are subjected to the same amount of heat. Which metal will undergo the largest change in temperature if neither metal goes through a phase change?

• the metal with the lower heat capacity

• the metal with the higher heat capacity

• the initial temperatures of both metals must be given

• both metals will undergo the same temperature change

Answer explanation

The lower a substance’s heat capacity, the less energy it takes to increase that substance’s temperature. The metal with the lower heat capacity will undergo the largest change in temperature as both metals are of the same mass and are not undergoing a phase change.

A system has constant volume (ΔV = 0) and the heat around the system increases by 45 J. What is the value of internal energy, ΔU = ΔE, of the system in Joules?

• - 45 J

• + 45 J

• +90 J

• -90J

Answer explanation

Since the system has constant volume (ΔV = 0) the term -PΔV = 0 and work is equal to zero. Thus, in the equation ΔU = ΔE = q + w, w = 0 and ΔU = ΔE = q. The internal energy is equal to the heat of the system. The surrounding heat increases, so the heat of the system decreases because heat is not created nor destroyed. Therefore, heat is taken away from the system making it exothermic and negative. The value of Internal Energy will be the negative value of the heat absorbed by the surroundings.

ΔU = ΔE = q + (-PΔV) = q + 0 = q

ΔU = ΔE  = - 45 J

A sample of an ideal gas having a volume of 0.70 L is allowed to expand against a constant opposing pressure of 0.50 atm until it has a volume of 1.0 L. What is the work for the system? Answer in units of J. (Hint: 1L•atm = 100 J)

• -15 J

• -0.15 J

• + 15 J

• + 1.5 J

Answer explanation

P = 0.50 atm               V₁ = 0.70 L                  V₂ = 1.0 L

In this case, the P is the constant opposing pressure of 0.50 atm. ΔV is the change in volume you observe, from 0.70 L to 1.0 L.

Plugging these values into the equation:

w = − PΔV = −(0.50 atm)(1.0 L − 0.70 L)

w = −(0.50 atm)(0.30 L) = - 0.15 L•atm

However, some conversion factors will have to be applied. To get to Joules, we use:

1L•atm = 100 J

Therefore, w = -15 J done by the system because ΔV is +.

A system had 150 kJ of work done on it and its internal energy increased by 60 kJ. How much energy did the system gain or lose as heat?

• The system lost 90 kJ of energy as heat

• The system gained 60 kJ of energy as heat

• The system gained 90 kJ of energy as heat.

• The system lost 210 kJ of energy as heat

Answer explanation

ΔU = q + w

q = ΔU − w = 60 kJ − (+150 kJ)

q = − 90 kJ       (Negative means the system lost energy as heat.)

When a stable, gaseous diatomic molecule spontaneously forms from individual gaseous atoms, for example 2 Cl _(g) --> Cl_{2 (g)},  what are the signs for ΔH ^o, ΔS ^o, and ΔG ^o?

• ΔH^o = −, ΔS^o = −, ΔG^o = +

• ΔH^o = −, ΔS^o = +, ΔG^o = −

• ΔH^o  = −, ΔS^o = −, ΔG^o = − 

• ΔH^o = +, ΔS^o = +, ΔG^o = −

Answer explanation

Forming bonds releases energy, so ΔH^o = −. Taking individual atoms and connecting them together into molecules lowers the entropy of the system, so ΔS^o = −. Finally, the diatomic molecule was formed spontaneously, meaning ΔG^o = −.

For the reaction 2 SO_{3 (g)} → 2 SO_{2 (g)} + O_{2 (g)}, ΔH^o_rxn = + 198 kJ/mol  at 298 K. Which statement is true for this reaction?

• The reaction will not be spontaneous at any temperatures

• ΔG^o_rxn will be negative at high temperatures.

• The reaction is driven by the enthalpy

• The reaction will not be spontaneous at high temperatures

Answer explanation

ΔG^o = ΔH^o− TΔS^o is used to predict spontaneity. ΔG^o is negative for a spontaneous reaction. ΔH^o is positive and T is always positive. For the reaction 2 mol gas → 3 mol gas. The more moles of gas, the higher the disorder, so ΔS^o is positive. Therefore, ΔG^o= (+) − T(+).

For ΔG^o to be negative, T must be large.