16. (JEE Main 2021 (Online) 25th February Morning Shift )

A diatomic gas, having $C_{p} = \frac{7}{2} R$ and $C_{v} = \frac{5}{2} R$ , is heated at constant pressure. The ratio dU : dQ : dW :

(A) 5 : 7 : 3

(B) 3 : 7 : 2

(C) 5 : 7 : 2

(D) 3 : 5 : 2

Correct answer is (C)

$d V = n \frac{5}{2} R Δ T$

$d Q = n \frac{7}{2} R Δ T$

$d W = d Q - d V$

$= n \frac{2}{2} R Δ T$

$∴$ $d V : d Q : d W$

$= n \frac{5}{2} R Δ T : n \frac{7}{2} R Δ T : n \frac{2}{2} R Δ T$

$= 5 : 7 : 2$

17. (JEE Main 2021 (Online) 1st September Evening Shift )

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 $\frac{x}{10}$ . Then the value of x (round off to the nearest integer) is ___________. (Given R = 8.31 J mol^{$-$ 1} K^{$-$ 1})

Correct answer is (25)

Given, the number of diatomic moles, n = 3 mol

The increase in temperature of the diatomic mole,

$Δ$ 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,

$Δ$ U = nC_v $Δ$ T .... (i)

where, $C_{v} = \frac{f}{2} R = \frac{5}{2} R$

Substituting the value in Eq. (i), we get

$Δ U = \frac{5 R}{2} n R Δ T$

Now, work done by the gas for isobaric process,

$W = p Δ V = n R Δ T$

The ratio of the change in internal energy to the work done by the gas,

$\frac{Δ U}{W} = \frac{\frac{5}{2} n R Δ T}{n R Δ T}$

$= \frac{Δ U}{W} = \frac{5}{2}$

Multiply and divide the above equation with 5, we get

$\frac{Δ U}{W} = \frac{5 \times 5}{2 \times 5} = \frac{25}{10}$

Comparing with given equation, $\frac{Δ U}{W} = \frac{x}{10}$

The value of the x = 25.

18. (JEE Main 2021 (Online) 26th February Evening Shift )

1 mole of rigid diatomic gas performs a work of $\frac{Q}{5}$ when heat Q is supplied to it. The molar heat capacity of the gas during this transformation is $\frac{x R}{8}$ . The value of x is _________. [R = universal gas constant]

Correct answer is (25)

From thermodynamics law:

$Q = Δ U + W$

$Q = Δ U + \frac{Q}{5}$

$Δ U = \frac{4 Q}{5}$

$n C_{v} Δ T = \frac{4}{5} n C Δ T$

$\frac{5}{4} C_{v} = C$

$C = \frac{5}{4} (\frac{f}{2}) R = \frac{5}{4} (\frac{5}{2}) R$

$C = \frac{25}{8} R$

$X = 25$

19. (JEE Main 2021 (Online) 26th February Evening Shift )

The volume V of a given mass of monoatomic gas changes with temperature T according to the relation $V = K T^{\frac{2}{3}}$ . The workdone when temperature changes by 90K will be xR. The value of x is _________. [R = universal gas constant]

Correct answer is (60)

We know that work done is

$W = \int P d V$ .... (1)

$\Rightarrow P = \frac{n R T}{V}$ .... (2)

$\Rightarrow W = \int \frac{n R T}{V} d v$ .... (3)

and given $V = K T^{2 / 3}$ .... (4)

$\Rightarrow W = \int \frac{n R T}{K T^{2 / 3}} . d v$ .... (5)

$\Rightarrow$ from (4) : $d v = \frac{2}{3} K T^{- 1 / 3} d T$

$\Rightarrow W = \int_{T_{1}}^{T_{2}} \frac{n R T}{K T^{2 / 3}} \frac{2}{3} K \frac{1}{T^{1 / 3}} d T$

$\Rightarrow W = \frac{2}{3} n R \times (T_{2} - T_{1})$ .... (6)

$\Rightarrow T_{2} - T_{1} = 90 K$ .... (7)

$\Rightarrow W = \frac{2}{3} n R \times 90$

$\Rightarrow W = 60 n R$

Assuming 1 mole of gas

n = 1

So, W = 60R

20. (JEE Main 2021 (Online) 25th February Morning Shift )

In a certain thermodynamical process, the pressure of a gas depends on its volume as kV³. The work done when the temperature changes from 100 $^{\circ}$ C to 300 $^{\circ}$ C will be ___________ nR, where n denotes number of moles of a gas.

Correct answer is (50)

$P = k v^{3}$

$\Rightarrow$ $p v^{- 3} = k$

$\Rightarrow$ $x = - 3$

$w = \frac{n R (T_{1} - T_{2})}{x - 1}$

$= \frac{n R (100 - 300)}{- 3 - 1}$

$= \frac{n R (- 200)}{- 4}$

$= 50$ nR

Thermodynamics