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Question: The specific heat of air at constant pressure is \(1.005{\text{ kJ/kg/K}}\) and the specific heat of...

The specific heat of air at constant pressure is 1.005 kJ/kg/K1.005{\text{ kJ/kg/K}} and the specific heat of air at constant volume is 0.718 kJ/kg/K0.718{\text{ kJ/kg/K}}. If the universal gas constant is 8.314 kJ/mole/K8.314{\text{ kJ/mole/K}}. Find the molecular weight of air.
(A)28.97(A)28.97
(B)24.6(B)24.6
(C)22.8(C)22.8
(D)19.6(D)19.6

Explanation

Solution

The universal gas constant is related to the difference between the specific heats (at constant pressure and at constant volume) of gas. Both of the specific heats are molar mass times of the specific heats for one gram of gas. From this, we will get that the specific gas constant of a gas or a mixture of gases is the ratio of the molar gas constant to the molar mass of the gas.

Formula used:
CP−CV=R{C_P} - {C_V} = R
The molar specific heat of air at constant volume CV=McV{C_V} = M{c_V}
The molar specific heat of air at constant pressure, CP=McP{C_P} = M{c_P}
RR is the molar gas constant.
cP−cV=r{c_P} - {c_V} = r
cP{c_P} is the specific heat at constant pressure for one gram of gas and cV{c_V} is the specific heat at constant volume for one gram of gas.
rr is the gas constant for one gram of gas.

Complete step by step solution:
At constant pressure and constant volume, the difference between the two specific heats of any gas, or we can say of air, is equal to the universal gas constant(R)(R).
CP−CV=R{C_P} - {C_V} = R
Now the gas constant can be represented as either a molar gas constant or as a gas constant for one gram of gas.
now,
CV=McV{C_V} = M{c_V}and CP=McP{C_P} = M{c_P}
CV{C_V} is the molar specific heat of air at constant volume and, CP{C_P} is the molar specific heat of air at constant pressure. MM is the molecular weight of air.
cP{c_P} is the specific heat at constant pressure for one gram of gas and cV{c_V} is the specific heat at constant volume for one gram of gas.
Also, cP−cV=r{c_P} - {c_V} = r
rr is the gas constant for one gram of gas. given,
so, from CP−CV=R{C_P} - {C_V} = R we can write
⇒McP−McV=R\Rightarrow M{c_P} - M{c_V} = R
⇒M(cP−cV)=R\Rightarrow M\left( {{c_P} - {c_V}} \right) = R
⇒cP−cV=RM\Rightarrow {c_P} - {c_V} = \dfrac{R}{M}
⇒r=RM\Rightarrow r = \dfrac{R}{M}
Given, cP=1.005 kJ/kg/K{c_P} = 1.005{\text{ kJ/kg/K}} and, cV=0.718 kJ/kg/K{c_V} = 0.718{\text{ kJ/kg/K}}
∴r=1.005−0.718=0.287kJ/kg/K\therefore r = 1.005 - 0.718 = 0.287{\text{kJ/kg/K}}
And, given R=8.314 kJ/mole/KR = 8.314{\text{ kJ/mole/K}}
⇒M=Rr\Rightarrow M = \dfrac{R}{r}
Putting the values of RR and rr, we get
⇒M=8.3140.287\Rightarrow M = \dfrac{{8.314}}{{0.287}}
⇒M=28.97\Rightarrow M = 28.97
Hence, the right answer is in Option: (A).

Note: The specific heat at constant pressure CP{C_P} is larger than the specific heat at constant volume CV{C_V} because of the addition of heat constant pressure, the material expands and works. As heat is added to gas at a fixed volume we get. QV=CVΔT=ΔU+W=ΔU{Q_V} = {C_V}\Delta T = \Delta U + W = \Delta U since no work is done.
Therefore,
dU=CV dTdU = {C_V}{\text{ }}dT and so, CV=dUdT{C_V} = \dfrac{{dU}}{{dT}}
When heat is added at constant pressure, we get
  QP=CPΔT=ΔU+ W=ΔU+PΔV\;{Q_P} = {C_P}\Delta T = \Delta U + {\text{ }}W = \Delta U + P\Delta V [ since, W=PΔVW = P\Delta V ]