Question

ecent communication with the inhabitants of Neptune has revealed that they have a Celsius-type temperature scale, but based on the melting point (0oN) and boiling point (100oN) of their most common substance, hydrogen. Further communications have revealed that the Neptunians know about perfect gas behavior and they find that, in the limit of zero pressure, the value of pV is 28 dm3atm at 0oN and 40 dm3atm at 100oN. What is the value of the absolute zero of temperature on their temperature scale

Answer 1

The absolute zero temperature is the smallest thermodynamic thermal limit, which is considered as a zero kelvin when the enthalpy, as well as the entropy of a cooled ideal gas, are of minimal value.

It is expressed by using the formula:

$\mathbf{T_{abs} = T_N + k}$

where;

$\mathbf{T_{abs } = absoulte \ temperature}$

$\mathbf{T_N = temperature \ at \Neptune \ Scale}$

k = factor to convert temperature from absolute scale to Neptunian scale

N:B: An absolute temperature = zero (0) kelvin

From the information given, Using Ideal gas since it is used in Neptune:

PV = nRT

where;

• P = pressure
• V = volume
• n = no of moles
• R = rate(constant)
• T = temperature

From the question, provided that the gas is the same and no new mass is added;

Then, we can equate the moles of gas in the given conditions as;

$\mathbf{\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}....\dfrac{P_nV_n}{T_n}}$

Using:

$\mathbf{\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}}$

where the value of;

P₁V₁ = 28 dm³ atm

T₁ = 0⁰ N

P₂V₂ = 40dm³ atm

T₂ =  100⁰ N

∴

$\mathbf{\dfrac{28}{(0+K)}=\dfrac{40}{(K+100)}}$

By cross multiply;

28(K+100) = 40(0+K)

28K + 2800  = 0 + 40K

40K - 28K = 2800

12 K = 2800

K = 2800/12

K = 233.33° N

Recall that:

$\mathbf{T_{abs} = T_N + k}$

$\mathbf{0= T_N+ 233.33 ^0 N}$

$\mathbf{ T_N= - 233.33 ^0 N}$

Therefore, we can conclude that the value of the absolute zero of temperature on their temperature scale is -233.33° N

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