Question

$C_v$ and $C_p$ denote the molar specific heat capacities of a gas at constant volume and constant pressure, respectively. Then
This question has multiple correct options
A) $C_p$$C_v$ is larger for diatomic ideal gas than for a monatomic ideal gas.
B) $C_p$ $+$ $C_v$ is larger for a diatomic ideal gas than for a monatomic ideal gas.
C) $\dfrac{{C_p}}{{C_v}}$ is larger for diatomic ideal gas than for a monatomic ideal gas.
D) ${C_p} \times{C_v}$ is larger for diatomic ideal gas than for a monatomic ideal gas.

Answer 1

Hint: For an ideal gas, $C_p$$-$$C_v$=R, where Cv and Cp denotes the molar heat capacities of an ideal gas at constant volume and constant pressure respectively and R is the gas constant and the value of R is 8.314 J/K..

Complete step-by-step answer:
Specific heat capacity is defined as the amount of heat required to raise the temperature of I kilogram of a substance by 1 kelvin(SI unit of Specific heat capacity $J\mathop {kg}\nolimits^{ - 1} \mathop k\nolimits^{ - 1} $
In case of free expansion under adiabatic conditions change in internal energy $\Delta $U=0. Then internal energy and temperature will remain constant.
(B) P$\alpha $\[\dfrac{1}{{\mathop v\nolimits^2 }}\]………..(1)
Therefore P$\mathop v\nolimits^2 $=constant
Then $T\alpha \dfrac{1}{V}$……….(2)
If volume is doubled then temperature will decrease as per the equation
(2) further, molar heat capacity in process, $P\mathop v\nolimits^X $=constant is
$C = \mathop C\nolimits_V = \dfrac{R}{{1 - X}}$
From equation (1), $x = 2$
Therefore, $C = \dfrac{3}{2}R = \dfrac{R}{{1 - 2}} = + \dfrac{R}{2}$
Since-, molar heat capacity is positive according to $Q = nc\Delta T$ will be negligible if $\Delta T$ is negative or gas loses heat if temperature is decreasing.
(C) $P\alpha \dfrac{1}{{\mathop V\nolimits^{4/3} }}$, $P\mathop V\nolimits^{4/3} $= constant
Therefore, $T\alpha \dfrac{1}{{\mathop V\nolimits^{1/3} }}$, further, with decrease in volume temperature will decrease.
Here, $X = \dfrac{4}{3}$, Then $C = \dfrac{3}{2}R + \dfrac{R}{{1 - \dfrac{4}{3}}} = - 1.5R$
As molar heat capacity is negative, $Q$ will be positive if $\Delta $T is negative or gas gains heat with decrease in temperature.

(D)$T\alpha PV$, in expansion from $V$ to $2V_1$, product of $P_V$ in increasing. Therefore, temperature will increase. Or $\Delta $U is positive. Further, in expansion work done is also positive.
Hence, $Q = W + \Delta U = $positive or gas gains heat.

So, the correct answers in options B and D.

Note: The ratio of these two varies as the atomicity of the gas changes. Also, their sum and product are not constant.