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

The ratio of work done by an ideal monoatomic gas to the heat supplied to it in an isobaric process is :

(A) $\frac{3}{5}$

(B) $\frac{3}{5}$

(C) $\frac{3}{2}$

(D) $\frac{2}{5}$

Correct answer is (D)

In an isobaric process,

Heat supplied, Q = n C_p $Δ$ T

Work done, w = nR $Δ$ T

$∴$    Ratio = $\frac{w}{Q}$ = $\frac{n R Δ T}{n C_{p} Δ T}$

=    $\frac{R}{\frac{5}{2} R}$

=    $\frac{2}{5}$

[C_p = $\frac{5}{2}$ R for monoatomic gas]

32. (JEE Main 2014 (Offline) )

One mole of a diatomic ideal gas undergoes a cyclic process $A B C$ as shown in figure. The process $B C$ is adiabatic. The temperatures at $A, B$ and $C$ are $400$ $K$ , $800$ $K$ and $600$ $K$ respectively. Choose the correct statement : JEE Main 2014 (Offline) Physics - Heat and Thermodynamics Question 298 English

(A) The change in internal energy in whole cyclic process is $250$ $R .$

(B) The change in internal energy in the process $C A$ is $700$ $R$ .

(C) The change in internal energy in the process $A B$ is - $350$ $R .$

(D) The change in internal energy in the process $B C$ is - $500$ $R .$

Correct answer is (D)

In cyclic process, change in total internal energy is zero.
$Δ U_{c y c l i c} = 0$
$Δ U_{B C} = n C_{v} Δ T = 1 \times \frac{5 R}{2} Δ T$

where, $C_{v} =$ molar specific heat at constant volume.
For $B C,$ $Δ T = - 200 K$
$∴$ $Δ U_{B C} = - 500 R$

33. (JEE Main 2013 (Offline) )

JEE Main 2013 (Offline) Physics - Heat and Thermodynamics Question 299 English

The above $p$ - $v$ diagram represents the thermodynamic cycle of an engine, operating with an ideal monatomic gas. The amount of heat, extracted from the source in a single cycle is

(A) $p_{0} v_{0}$

(B) $(\frac{13}{2}) p_{0} v_{0}$

(C) $(\frac{11}{2}) p_{0} v_{0}$

(D) $4 p_{0} v_{0}$

Correct answer is (B)

Along path DA, volume is constant.

Hence, $Δ$ Q_DA = nC_v $Δ$ T = nC_v(T_A – T_D)

$∴$ $Δ$ Q_DA = $n (\frac{3}{2} R) [\frac{2 p_{0} v_{0}}{n R} - \frac{p_{0} v_{0}}{n R}] = \frac{3}{2} p_{0} v_{0}$

Along the path AB, pressure is constant.

Hence $Δ$ Q_AB = nC_p $Δ$ T = nC_p(T_B – T_A)

$∴$ $Δ$ Q_AB = $n (\frac{5}{2} R) [\frac{2 p_{0} 2 v_{0}}{n R} - \frac{2 p_{0} v_{0}}{n R}] = \frac{10}{2} p_{0} v_{0}$

$∴$ The amount of heat extracted from the source in a single cycle is

$Δ$ Q = $Δ$ Q_DA + $Δ$ Q_AB

$= \frac{3}{2} p_{0} v_{0} + \frac{10}{2} p_{0} v_{0}$ = $(\frac{13}{2}) p_{0} v_{0}$

34. (AIEEE 2011 )

$100 g$ of water is heated from $30^{\circ} C$ to $50^{\circ} C$ . Ignoring the slight expansion of the water, the change in its internal energy is (specific heat of water is $4184$ $J / k g / K$ ):

(A) $8.4$ $k J$

(B) $84$ $k J$

(C) $2.1$ $k J$

(D) $4.2$ $k J$

Correct answer is (A)

$Δ U = Δ Q = m c Δ T$
$= 100 \times 10^{- 3} \times 4184 (50 - 30) \approx 8.4 k J$

35. (AIEEE 2009 )

Two moles of helium gas are taken over the cycle $A B C D,$ as shown in the $P$ - $T$ diagram. AIEEE 2009 Physics - Heat and Thermodynamics Question 303 English

The work done on the gas in taking it from $D$ to $A$ is :

(A) $+ 414$ $R$

(B) $- 690$ $R$

(C) $+ 690$ $R$

(D) $- 414$ $R$

Correct answer is (A)

Work done by the system in the isothermal process
$D A$ is $W = 2.303 n R T \log_{10} \frac{P_{D}}{P_{A}}$
$= 2.303 \times 2 R \times 300 \log_{10} \frac{1 \times 10^{5}}{2 \times 10^{5}} = - 414 R .$
Therefore work done on the gas is $+ 414 R .$