If the ratio of the specific heats of steam is 1.33 and R = 8312J/k mole K find the molar heat capacities of steam at constant pressure and constant volume.
A) $33.5kJ/k mole,25.19kJ/kgK$
B) $25.19kJ/k,33.5kJ/kg K$
C) $18.82 kJ/k,10.82kJ/k\,mole$
D) None of these
Answer
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Hint: For any gas, the ratio of specific heat at constant pressure and that at constant volume is a fixed quantity that depends on the molecularity of that gas. Again the difference between these two specific heats is the same as the universal gas constant. So, using these two relations both the specific heats can be found.
Formula used:
The ratio of specific heat at constant pressure and specific heat at constant volume of a gas is given as:
$\gamma = \dfrac{{{C_P}}}{{{C_V}}}$ …………..(1)
Where,
$\gamma $is the specific heat ratio,
${C_P}$is the specific heat of a gas at constant pressure,
${C_V}$ is the specific heat of a gas at constant volume.
The relation between two specific heats is given by:
${C_P} - {C_V} = R$ …………………. (2)
Where,
R is the universal gas constant.
Complete step by step answer:
Given, The ratio of specific heats is $\gamma = 1.33$.
The value of the universal gas constant is $R=8312J/k \, mole\, K= 8.132 kJ/k\, mole\, K$
To find: Molar heat capacities of the steam at constant pressure and constant volume i.e. ${C_P}$and ${C_V}$respectively.
Step 1:
First, divide both sides of eq.(2) with ${C_V}$and substitute the relation from eq.(1) to get an expression for ${C_V}$ as:
$ \dfrac{C_P}{C_V}-1=\dfrac{R}{C_V}$
$\Rightarrow \dfrac{R}{C_V}=\gamma -1$
$\Rightarrow \dfrac{C_V}{R}=\dfrac{1}{\gamma-1}$
$\Rightarrow {C_V}=\dfrac{R}{\gamma - 1}$ ………….(3)
Step 2:
Substitute the values of $\gamma $ and R in eq.(3) to get the value of ${C_V}$ as:
${C_V} = \dfrac{8.312kJ/k\,mole.K}{1.33-1}$
Step 3:
Now, substitute the value of R and ${C_V}$from eq.(4) into eq.(2) to get the value of ${C_P}$ as,
$ {C_P} = R + {C_V} $
$\Rightarrow {C_P} = 8.312{kJ/k\, mole K}+25.19{kJ/k\, mole K}$
$\therefore {C_P}=33.5{kJ/k\, mole K}$ ………(4)
Molar heat capacities of steam at constant pressure and constant volume are 33.5kJ/k mole.K, 25.19kJ/kgK respectively. So, option (A) is correct.
Note:
This problem has a very tricky solution. You can solve it without any calculation just by observing the options. Focus on eq.(2). The difference between ${C_P}$ and ${C_V}$ is always R and ${C_P}$ is greater. Now, notice that among the given options only option (a) satisfies it. To discard option (d) just take their ratio and you will approximately get 1.33. Hence, option (a) is the correct choice.
Formula used:
The ratio of specific heat at constant pressure and specific heat at constant volume of a gas is given as:
$\gamma = \dfrac{{{C_P}}}{{{C_V}}}$ …………..(1)
Where,
$\gamma $is the specific heat ratio,
${C_P}$is the specific heat of a gas at constant pressure,
${C_V}$ is the specific heat of a gas at constant volume.
The relation between two specific heats is given by:
${C_P} - {C_V} = R$ …………………. (2)
Where,
R is the universal gas constant.
Complete step by step answer:
Given, The ratio of specific heats is $\gamma = 1.33$.
The value of the universal gas constant is $R=8312J/k \, mole\, K= 8.132 kJ/k\, mole\, K$
To find: Molar heat capacities of the steam at constant pressure and constant volume i.e. ${C_P}$and ${C_V}$respectively.
Step 1:
First, divide both sides of eq.(2) with ${C_V}$and substitute the relation from eq.(1) to get an expression for ${C_V}$ as:
$ \dfrac{C_P}{C_V}-1=\dfrac{R}{C_V}$
$\Rightarrow \dfrac{R}{C_V}=\gamma -1$
$\Rightarrow \dfrac{C_V}{R}=\dfrac{1}{\gamma-1}$
$\Rightarrow {C_V}=\dfrac{R}{\gamma - 1}$ ………….(3)
Step 2:
Substitute the values of $\gamma $ and R in eq.(3) to get the value of ${C_V}$ as:
${C_V} = \dfrac{8.312kJ/k\,mole.K}{1.33-1}$
Step 3:
Now, substitute the value of R and ${C_V}$from eq.(4) into eq.(2) to get the value of ${C_P}$ as,
$ {C_P} = R + {C_V} $
$\Rightarrow {C_P} = 8.312{kJ/k\, mole K}+25.19{kJ/k\, mole K}$
$\therefore {C_P}=33.5{kJ/k\, mole K}$ ………(4)
Molar heat capacities of steam at constant pressure and constant volume are 33.5kJ/k mole.K, 25.19kJ/kgK respectively. So, option (A) is correct.
Note:
This problem has a very tricky solution. You can solve it without any calculation just by observing the options. Focus on eq.(2). The difference between ${C_P}$ and ${C_V}$ is always R and ${C_P}$ is greater. Now, notice that among the given options only option (a) satisfies it. To discard option (d) just take their ratio and you will approximately get 1.33. Hence, option (a) is the correct choice.
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