Question 545: Thermodynamics & KTG

Question

The temperature of 3.00 mol of an ideal diatomic gas is increased by 40.0$$^\circ$$C without changing the pressure of the gas. The molecules in the gas rotate but do not oscillate. If the ratio of change in internal energy of the gas to the amount of workdone by the gas is $${x \over {10}}$$. Then the value of x (round off to the nearest integer) is ___________. (Given R = 8.31 J mol
$$-$$1
K
$$-$$1
)

Accepted Answer

Given, the number of diatomic moles, n = 3 mol
The increase in temperature of the diatomic mole,
$$\Delta$$T = 40$$^\circ$$C
Now, the degree of freedom
f = linear + rotational + no oscillation
f = 3 + 2 + 0 $$\Rightarrow$$ f = 5
Change in internal energy,
$$\Delta$$U = nC
v
$$\Delta$$T .... (i)
where, $${C_v} = {f \over 2}R = {5 \over 2}R$$
Substituting the value in Eq. (i), we get
$$\Delta U = {{5R} \over 2}nR\Delta T$$
Now, work done by the gas for isobaric process,
$$W = p\Delta V = nR\Delta T$$
The ratio of the change in internal energy to the work done by the gas,
$${{\Delta U} \over W} = {{{5 \over 2}nR\Delta T} \over {nR\Delta T}}$$
$$ = {{\Delta U} \over W} = {5 \over 2}$$
Multiply and divide the above equation with 5, we get
$${{\Delta U} \over W} = {{5 	imes 5} \over {2 	imes 5}} = {{25} \over {10}}$$
Comparing with given equation, $${{\Delta U} \over W} = {x \over {10}}$$
The value of the x = 25.

Thermodynamics & KTG

Integer / Numeric Answer Type