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At constant volume the specific heat of a gas is $\frac{3 R}{2}$ , then the value of $' γ'$ will be

$\frac{3}{2}$

$\frac{5}{2}$

$\frac{5}{3}$

None of the above

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Explanation

$3 . C_{V} = \frac{R}{(γ - 1)} \Rightarrow γ = 1 + \frac{R}{C_{V}} = 1 + \frac{R}{\frac{3}{2} R} = \frac{5}{3}$

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The relation between two specific heats of a gas is

$C_{P} - C_{V} = \frac{R}{J}$

$C_{V} - C_{P} = \frac{R}{J}$

$C_{P} - C_{V} = J$

$C_{V} - C_{P} = J$

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Explanation

1. When $C_{P}$ and $C_{V}$ are given with calorie and R with Joule then $C_{P} - C_{V} = R / J .$

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The specific heat of an ideal gas is

Proportional to $T$

Proportional to $T^{2}$

Proportional to $T^{3}$

Independent of $T$

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Explanation

4. According to the equilibrium theorem, the molar heat capacities should be independent of

temperature. However, variations in $C_{V}$ and $C_{P}$ are observed as the temperature changes. At very

high temperatures, vibrations are also important and that affects the values of $C_{V}$ and $C_{P}$ for

diatomic and polyatomic gases. Here in this question according to given information option 4 may be

correct answer.

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Molar specific heat at constant volume is $C_{V}$ for a monoatomic gas is

$\frac{3}{2} R$

$\frac{5}{2} R$

$3 R$

$2 R$

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Explanation

$1 . C_{V} = \frac{f}{2} R$ For monoatomic gas $f = 3 \Rightarrow {(C_{V})}_{mono} = \frac{3}{2} R .$

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The following sets of values for $C_{V}$ and $C_{P}$ of a gas has been reported by different students. The units are cal/gm-mole-K. Which of these sets is most reliable

$C_{V} = 3, C_{P} = 5$

$C_{V} = 4, C_{P} = 6$

$C_{V} = 3, C_{P} = 2$

$C_{V} = 3, C_{P} = 4.2$

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Explanation

1. $C_{P} - C_{V} = R = 2 . \frac{cal}{gm - mol - K}$ which is correct for option 1 and 2. Further the ratio $\frac{C_{P}}{C_{V}} (= γ)$ should

equal to some standard value corresponding to that of either, mono, di, or triatomic gases. From this

point of view option 1 is correct because ${(\frac{C_{P}}{C_{V}})}_{mono} = \frac{5}{3}$

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The specific heats at constant pressure is greater than that of the same gas at constant volume because

At constant pressure work is done in expanding the gas

At constant volume work is done in expanding the gas

The molecular attraction increases more at constant pressure

The molecular vibration increases more at constant pressure

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Explanation

1. $C_{P} > C_{V}$

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The specific heat of a gas

Has only two values $C_{P}$ and $C_{V}$

Has a unique value at a given temperature

Can have any value between 0 and $\infty$

Depends upon the mass of the gas

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Explanation

$3 . C_{isothermal} = \infty and C_{adiabatic} = 0$

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For a gas if $γ = 1.4$ , then atomicity, $C_{P}$ and $C_{V}$ of the gas are respectively

Monoatomic, $\frac{5}{2} R, \frac{3}{2} R$

Monoatomic, $\frac{7}{2} R, \frac{5}{2} R$

Diatomic, $\frac{7}{2} R, \frac{5}{2} R$

Triatomic, $\frac{7}{2} R, \frac{5}{2} R$

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Explanation

$3 . γ = 1 + \frac{2}{f} \Rightarrow 1.4 = 1 + \frac{2}{f} \Rightarrow Degree of freedom f = 5 \Rightarrow Degree of freedom of diatomic gas is 5 and it' s C_{P} = \frac{7}{2} R and C_{V} = \frac{5}{2} R$

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For a gas the difference between the two specific heats is 4150 J/kg K. What is the specific heats at constant volume of gas if the ratio of specific heat is 1.4

8475 J/Kg K

5186 J/Kg K

1600 J/Kg K

10375 J/Kg K

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Explanation

$4 . C_{P} - C_{V} = R = 4150 \frac{J}{kg - K} and \frac{C_{P}}{C_{V}} = γ = 1.4 \Rightarrow C_{V} = \frac{R}{(γ - 1)} = \frac{4150}{(1.4 - 1)} = 10375 J / kg - K$

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The specific heat of 1 mole of an ideal gas at constant pressure $(C_{P})$ and at constant volume $(C_{V})$ which is correct

$C_{P}$ of hydrogen gas is $\frac{5}{2} R$

$C_{V}$ of hydrogen gas is $\frac{7}{2} R$

$H_{2}$ has very small values of $C_{P}$ and $C_{V}$

$C_{P} - C_{V} =$ 1.99 cal/mole-K for $H_{2}$

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Explanation

4. For any gas $C_{P} - C_{V} = 1.99 \approx 2 \frac{cal}{mol - K}$

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