QJEEM21Ph1b07: JEE Main 2021 Physics - Class 11 JEEM21Ph1b07 Solution: The temperature of 3.00 mol of an ideal diatomic gas is…

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

${{\Delta U} \over W} = {{5 \times 5} \over {2 \times 5}} = {{25} \over {10}}$

Comparing with given equation, ${{\Delta U} \over W} = {x \over {10}}$

The value of the x = 25.