26. (JEE Main 2019 (Online) 10th April Morning Slot )

n moles of an ideal gas with constant volume heat capcity C_V undergo an isobaric expansion by certain volume. The ratio of the work done in the process, to the heat supplied is :

(A) $\frac{n R}{C_{V} - n R}$

(B) $\frac{4 n R}{C_{V} - n R}$

(C) $\frac{4 n R}{C_{V} + n R}$

(D) $\frac{n R}{C_{V} + n R}$

Correct answer is (D)

w = nR $Δ$ T
$Δ$ H = (Cv + nR) $Δ$ T
$\frac{ω}{Δ H} = \frac{n R}{C_{v} + n R}$

27. (JEE Main 2019 (Online) 9th April Morning Slot )

Following figure shows two processes A and B for a gas. If $Δ$ Q_A and $Δ$ Q_B are the amount of heat absorbed by the system in two cases, and $Δ$ U_A and $Δ$ U_B are changes in internal energies, respectively, then : JEE Main 2019 (Online) 9th April Morning Slot Physics - Heat and Thermodynamics Question 243 English

(A) $Δ$ Q_A > $Δ$ Q_B ; $Δ$ U_A > $Δ$ U_B

(B) $Δ$ Q_A < $Δ$ Q_B ; $Δ$ U_A < $Δ$ U_B

(C) $Δ$ Q_A > $Δ$ Q_B ; $Δ$ U_A = $Δ$ U_B

(D) $Δ$ Q_A= $Δ$ Q_B ; $Δ$ U_A = $Δ$ U_B

Correct answer is (C)

Initial and final states for both the processes are same,
$∴$ $Δ$ U_A = $Δ$ U_B

Work done during process A is greater than in process B. Because area is more By First law of thermodynamics
$Δ$ Q = $Δ$ U + W
$\Rightarrow$ $Δ$ Q_A > $Δ$ Q_B

28. (JEE Main 2019 (Online) 12th January Morning Slot )

For the given cyclic process CAB as shown for a gas, the work done is :

JEE Main 2019 (Online) 12th January Morning Slot Physics - Heat and Thermodynamics Question 251 English

(A) 1 J

(B) 10 J

(C) 5 J

(D) 30 J

Correct answer is (B)

Since P $-$ V indicator diagram is given, so work done by gas is area under the cyclic diagram.

$∴$ $Δ$ W = Work done by gas = $\frac{1}{2}$ $\times$ 4 $\times$ 5 J

= 10 J

29. (JEE Main 2019 (Online) 9th January Morning Slot )

A gas can be taken from A to B via two different processes ACB and ADB.

JEE Main 2019 (Online) 9th January Morning Slot Physics - Heat and Thermodynamics Question 269 English
When path ACB is used 60 J of heat flows into the system and 30 J of work is done by the system. If path ADB is used work done by the system is 10 J. The heat Flow into the system in path ADB is :

(A) 40 J

(B) 80 J

(C) 100 J

(D) 20 J

Correct answer is (A)

Using law of thermodynamics in path ACB,

$Δ$ Q_ACB = $Δ$ U_ACB + $Δ$ W_ACB

$\Rightarrow$    60 = $Δ$ U_ACB + 30

$\Rightarrow$     $Δ$ U_ACB = 30 J

As value of $Δ$ U is path independent,

$∴$     $Δ$ U_ACB = $Δ$ U_ADB = 30 J

$∴$    In path ADB,

$Δ$ Q_ADB = $Δ$ U_ADB + $Δ$ W_ADB

$\Rightarrow$ $Δ$ Q_ADB = 30 + 10 = 40 J

30. (JEE Main 2016 (Online) 9th April Morning Slot )

200 g water is heated from 40^oC to 60^oC. Ignoring the slight expansion of water, the change in its internal energy is close to (Given specific heat of water = 4184 J/kg/K) :

(A) 8.4 kJ

(B) 4.2 kJ

(C) 16.7 kJ

(D) 167.4 kJ

Correct answer is (C)

According to the first law of thermodynamics,

Q = $Δ$ u + w

For isochoric process Q = $Δ$ U = ms $Δ$ t

$Δ$ T = (233 $-$ 213) = 20 k

$∴$ $Δ$ u = 0.2 $\times$ 4184 $\times$ 20 = 16.7 kJ