
The molar specific heat of the process $V\alpha {T^4}$for $C{H_4}$ gas at room temperature is:
(A) 4R
(B) 7R
(C) 3R
(D) 8R
Answer
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Hint To answer this question at first we have to develop the expression between the temperature and pressure. Since P and V are both constant in the given process, we have to find the value of $\Delta Q$ from the expression. Then we have to find the value of C or molar specific heat. This will give us the answer to the required question.
Complete step by step answer
We know that $V{T^{ - 4}}$= constant.
Hence we can write that:
$V.{(PV)^{ - 4}}$= constant
$ \Rightarrow {P^{ - 4}}{V^{ - 3}}$= constant
Therefore, we can write that:
$P{V^{3/4}}$= constant.
Hence now we can write that:
$w = \dfrac{{nR\Delta T}}{{1 - \dfrac{3}{4}}} = 4nRT$
After the above expression we can write that:
$u = \dfrac{1}{2}nR\Delta T$
Now we have to find the value of $\Delta Q$. So we can write the expression as:
$\Delta Q = n(\dfrac{1}{2} + 4)R\Delta T$
Therefore, we have to find the value of the molar specific heat or C, which is given as:
$C = \left( {\dfrac{1}{2} + 4} \right)R$
Hence the value of f is 8.
So as the value of f is 8, we have to put the value in the equation involving the term R, to get the answer.
So we can say that C = 8R
Hence we can say that the molar specific heat of the process $V\alpha {T^4}$ for $C{H_4}$ gas at room temperature is 8R.
Therefore, the correct answer is option D.
Note For better understanding we should be knowing the meaning of molar specific heat. So by molar specific heat of a solid or a liquid is defined as the heat that will be provided to the heat so as to raise the temperature of one mole of the solid or liquid by 1 Kelvin or 1 degree Celsius. The unit of molar specific heat is $Jmo{l^{ - 1}}{K^{ - 1}}$.
We should also remember that the molar heat capacity generally will increase along with the molar mass. The molar specific heat will vary with both the temperature and the pressure, and the result will be different for each of the different states of matter.
Complete step by step answer
We know that $V{T^{ - 4}}$= constant.
Hence we can write that:
$V.{(PV)^{ - 4}}$= constant
$ \Rightarrow {P^{ - 4}}{V^{ - 3}}$= constant
Therefore, we can write that:
$P{V^{3/4}}$= constant.
Hence now we can write that:
$w = \dfrac{{nR\Delta T}}{{1 - \dfrac{3}{4}}} = 4nRT$
After the above expression we can write that:
$u = \dfrac{1}{2}nR\Delta T$
Now we have to find the value of $\Delta Q$. So we can write the expression as:
$\Delta Q = n(\dfrac{1}{2} + 4)R\Delta T$
Therefore, we have to find the value of the molar specific heat or C, which is given as:
$C = \left( {\dfrac{1}{2} + 4} \right)R$
Hence the value of f is 8.
So as the value of f is 8, we have to put the value in the equation involving the term R, to get the answer.
So we can say that C = 8R
Hence we can say that the molar specific heat of the process $V\alpha {T^4}$ for $C{H_4}$ gas at room temperature is 8R.
Therefore, the correct answer is option D.
Note For better understanding we should be knowing the meaning of molar specific heat. So by molar specific heat of a solid or a liquid is defined as the heat that will be provided to the heat so as to raise the temperature of one mole of the solid or liquid by 1 Kelvin or 1 degree Celsius. The unit of molar specific heat is $Jmo{l^{ - 1}}{K^{ - 1}}$.
We should also remember that the molar heat capacity generally will increase along with the molar mass. The molar specific heat will vary with both the temperature and the pressure, and the result will be different for each of the different states of matter.
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