Thermal Physics - Isothermal

Suppose one mole of an ideal gas undergoes the reversible cycle ABCA shown in the P-V diagram above, where AB is an isotherm. The molar heat capacities are C_p at constant pressure and C_v at constant volume. The net heat added to the gas during the cycle is equal to

A. RT_h (V₂/V₁)
B. −C_p(T_h − T_c)
C. C_p(T_h − T_c)
D. RT_h ln (V₂/V₁) − C_p(T_h − T_c)
E. RT_h ln (V₂/V₁) − R(T_h − T_c)

(GR9677 #15)

Solution:

AB Isotherm → T Constant = T_h

Ideal Gas: PV = nRT 
P = nRT / V

n = 1 mole,

W_AB = _V1∫^V₂ P dV = RT_h V1∫^V₂ (1/V) dV  = RT_h ln (V₂/V₁)

BC Isobaric → P constant = P₂

W_BC = _V2∫^V₁ P dV = P₂ (V₁− V₂) =  P₂V₁ −  P₂V₂

From the diagram:

P₂V₁ = nRT_c 

P₂V₂ = nRT_h

W_BC = nR(T_c − T_h)  = R(T_c − T_h)

W_CA = 0  since V constant

Total W = W_AB + W_BC = RT_h ln (V₂/V₁) + R(T_c − T_h)

or

W = RT_h ln (V₂/V₁) − R(T_h − T_c)

Answer: E

Thermal Physics - Heat Capacity

A Classical model of a diatomic molecule is a springy dumbbell, as shown above, where the dumbbell is free to rotate about axes perpendicular to the spring. In the limit of high temperature, what is the specific heat per mole at constant volume?

A. 3/2 R
B. 5/2 R
C. 7/2 R
D. 9/2 R
E. 11/2 R

(GR9277 #15) 

Solution:

C_V  for diatomic gas: C_V  = C_Vtrans  + C_Vrot  + C_Vvib

At high T, all components contribute and each degree of freedom (DoF) contributes ½R to C_V .

DoF trans = 3 (since it's 3 dimension)
DoF rot = 2 (diatomic atom)
DoF vib = 2 (1 for kinetic energy + 1 potential energy)
→ 3 + 2 + 2 = 7 → 7 DoF contributes ⁷/₂R

Answer: C

Thermal Physics - Probability

A sample of N atoms of helium gas is confined in a 1.0 cubic meter volume. The probability that none of the helium atoms is in a 10⁻⁶ cubic meter volume of the container is

A. 0
B. (10⁻⁶)^N
C. (1 − 10⁻⁶)^N
D. 1 − (10⁻⁶)^N
E. 1

(GR8677 #15)

Solution:

Total Probability: P = P₁ + P₂ = 1
P₁ = The probability that one atom is in a 10⁻⁶ cubic meter volume of the container
P₂ =  The probability that none of the helium atoms is in a 10⁻⁶ cubic meter volume of the container
→ P₂ = 1 − P₁

P₁ = 1/n
n = number of 10⁻⁶ m³ cubes in the 1 m³ volume
10⁻⁶ × n = 1 → n = 1/10⁻⁶ = 10⁶ cubes
→ P₁ = 1/n = 1/10⁶ = 10⁻⁶
→ P₂ = 1 − P₁ = 1 − 10⁻⁶
For N atoms: P₂ = (1 − 10⁻⁶)^N

Answer: C

Lab Methods - Precision

Five classes of students measure the height of a building. Each classes uses a different method and each measures the height many different times. The data for each class are plotted below. Which class made the most precise measurement?

(GR0177 #15)

Solution:

The accuracy is how close the peak is to the reference value.

The precision is how narrow the peak is.

So we look for the graph with the narrowest peak.

Answer: A