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# The heat capacity at constant volume of a sample of a monatomic gas is 35J/k. Find the molar heat capacity at constant pressure for 1 mol of gas.

Last updated date: 13th Jun 2024
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Hint: Firstly, we will write the 1st law of thermodynamics $dQ = dU + dW$ . Then we will see that constant volume change in heat becomes equal to the change in internal energy when volume remains constant. After that we will write the value of molar specific heat at constant volume and in constant pressure. Then we will find out the ratio between value of molar specific heat at constant volume and in constant pressure. Then we will calculate the molar heat capacity at constant pressure for 1 mol of gas after putting all the values.

Complete step by step solution
We know that the first law of thermodynamics says that when system takes some amount of heat from its surroundings, it is used up in two ways: A part of it increases the internal energy of the system and the remaining part is conserved into some external work done by the system that is heat absorbed= rise in internal energy external work done.
So from the first law of thermodynamics we can write if a small amount of heat dQ changes the internal energy of the system by dU and an external work dW is done then, $dQ = dU + dW$
At constant volume, $dV = 0$ so $dW = pdV = 0$ .
So, $dQ = dU$
So, we can write that the heat gained or lost at constant volume for a temperature change dT of 1 mol of a gas is $dQ = {C_v}dT$
Here, ${C_v}$ is the molar specific heat at constant volume
${C_v} = \dfrac{{dQ}}{{dT}} = \dfrac{{dU}}{{dT}}$
The total internal energy of 1 mol of an ideal gas is $U = \dfrac{3}{2}RT$
So, ${C_v} = \dfrac{{dU}}{{dT}} = \dfrac{3}{2}R....(i)$
The molar specific heat at constant pressure is ${C_p} = {C_v} + R = \dfrac{3}{2}R + R = \dfrac{5}{2}R.....(ii)$
The ratio between the two specific heats is called heat capacity ratio
$\gamma = \dfrac{{{C_p}}}{{{C_V}}} = \dfrac{{\dfrac{5}{2}R}}{{\dfrac{3}{2}R}} = \dfrac{5}{3}....(iii)$
$\Rightarrow {C_P} = {C_V} \times \gamma \\ \Rightarrow {C_P} = \dfrac{5}{3}{C_V} = 58.33\dfrac{J}{{mol - K}} \\$

Hence the required solution is 58.33J/mol-K

Additional Information: The average molecular kinetic energy of any substance is equally shared among the degrees of freedom; the average kinetic energy of a single molecule associated with each degree of freedom is $\dfrac{1}{2}kT$ , here T=absolute temperature and k= Boltzmann constant. The number of degrees of freedom of an ideal gas molecule is 3. So, from the equipartition of energy we can write the average kinetic energy of a molecule $= 3 \times \dfrac{1}{2}kT = \dfrac{3}{2}kT$ . The molecule has no potential energy so average total energy $e = \dfrac{3}{2}kT$ and the total internal energy of 1 mol of an ideal gas is $U = \dfrac{3}{2}RT$

Note: The molecules of monoatomic gas can move in any direction in space so it has three independent motions and hence 3 degrees of freedom. So one may put the wrong value of degrees of freedom for a monatomic gas. Especially every 3 dimensional motion has 3 degrees of motion.
Alternate method:
The minimum number of independent coordinates necessary to specify the instantaneous position of a moving body is called the degree of freedom of the body.
We know that for monoatomic gas, degree of freedom is 3
From equation (i) we can write ${C_v} = \dfrac{f}{2}R[\because f = 3]$
And ${C_P} = \dfrac{f}{2}R + R = \dfrac{{f + 2}}{2}R$
So, from equation (iii) we can write $\gamma = \dfrac{{{C_p}}}{{{C_V}}} = \dfrac{{f + 2}}{f} = 1 + \dfrac{2}{f}$
Given that ${C_V} = 35\dfrac{J}{k}$
So, we can write, ${C_P} = \gamma {C_V} = \dfrac{{f + 2}}{f}{C_V} = \dfrac{5}{3}{C_V} = 58.33\dfrac{J}{{mol - K}}$