physics ncert solutions class 11th

This is a short answer type question as classified in NCERT Exemplar

During driving temperature of the gas increases while volume remains constant. So according charle's law, at constant volume V.

Pressure is directly proportional to temperature. Therefore pressure of gas increases.

This is a short answer type question as classified in NCERT Exemplar

Yes during adiabatic compression the temperature of a gas increases while no heat is

In adiabatic compression dQ=0

From the first law of thermodynamics dU= dQ-dW

dU=-dW

in compression work is done on the gas i.e work done is negative

dU=positive

hence internal energy of the gas increases due to which its temperature increases.

This is a short answer type question as classified in NCERT Exemplar

If a refrigerator's doors is kept open, then room will become hot, because amount of heat removed would be less than the amount of heat released in the room.

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For path1

Heat Q₁= 1000J

Work done =W₁

For path 2

Work done W₂= W₁-100

As change in internal energy is same

dU=Q₁-W₁=Q₂-W₂

1000-W₁=Q₂-W₁+100

Q₂= 1000-100= 900J

This is a short answer type question as classified in NCERT Exemplar

Yes this is possible when the entire heat supplied to the system is utilised in expansion.

So its working against the surroundings.

This is a long answer type question as classified in NCERT Exemplar

(a) Initially the piston is in equilibrium P_i=P_a

(b) On supplying heat , the gas expands from V_o to V_i

so increase in volume of the gas =V_i-V_o

as the piston is of unit cross sectional area hence extension in the spring

x= $\frac{V_i-V_0}{area}=V_i-V_0$

force exerted by the spring on the piston= F= kx= K(V_i - V_o)

hence final pressure =P_f =P_a +kx

= P_a+K $*$ ( $V_i-V_0$ )

(c) From first law of thermodynamics

dQ=dU+dW

dU=C_v(T-T_o) = C_v(T-T_o)

T= $\frac{P_f{V}_1}{R}=\left[\frac{P_a+K\left(V_i-V_0\right)}{R}\right]\frac{V_1}{R}$

Work done by the gas =pdV+ increase in PE of the spring

= P_a(V₁-V_o) + $\frac{1}{2}k$ x²

dQ=dU+dW

= C_v(T-T_o)+P_a(V-V_o)+ $\frac{1}{2}k$ x²

= C_v(T-T_o)+P_a(v-V_o)+1/2 ( $V_i-V_0$ )²

This is a long answer type question as classified in NCERT Exemplar

Slope of the curve = f(V) , where V is the volume

Slope of P = f(V) curve at ((P_o, V₀ )= f(V_o)

Slope of adiabatic at (P_o, V₀ )= k(-Y)V_o^-1-Y =-YP_o/V_o

Now heat absorbed in the process P= f(V)

dQ=dU+dW= nC_vdT+pdV

pV=nRT

T= pV/nR

T= $\frac{1}{nR}\left[f\left(V\right)+V{f}^{\text{'}}\left(V\right)dV\right]$

$\frac{dQ}{dV}=$ nC_{v $\frac{dT}{dV}+p\frac{dV}{dV}=n{C}_V\frac{dT}{dV}+P$}

$\frac{C_V}{R}\left[f\right(V_o+V_0{f}^{\text{'}}\left(V\right)+f\left(V\right)]$

After solving we get

${\left[\frac{dQ}{dV}\right]}_{v=vo}=$ $\left[\frac{1}{Y-1}+1\right]f\left(V_o\right)+\frac{V_o{f}^{\text{'}{V}_o}}{Y-1}$

= $\frac{Y}{Y-1}{P}_o+\frac{V_o}{Y-1}{f}^{\text{'}}\left(V_o\right)$

Heat is absorbed where dQ/dV>0  when gas expands

Hence YP_o+V_of'(V_o)>0  or f'(V_o)>(-Y $\frac{P_o}{V_o}$ )

This is a long answer type question as classified in NCERT Exemplar

(a) For process AB

Volume is constant , hence work done dW=0

dQ=dU+dW=dU+0=dU

 = nC_vdT= nC_v(T_B-T_A)

 = $\frac{3}{2}R\left(T_B-T_A\right)$

= $\frac{3}{2}\left(R{T}_B-R{T}_A\right)=\frac{3}{2}\left(P_B{V}_B-P_A{V}_A\right)$

Heat exchanged = $\frac{3}{2}\left(P_B{V}_B-P_A{V}_A\right)$

(b) For process BC , p =constant

dQ= dU+dW  = $\frac{3}{2}R\left(T_C-T_B\right)+P_B(V_C-V_B)$

heat exchanged = $\frac{5}{2}{P}_B\left(V_C{-V}_A\right)$

(c) For process CD , because CD is adiabatic , dQ= heat exchanged =0

(d) DA involves compression of gas from V_D to V_A at constant pressure P_A

heat transferred as similar way as BC₁

hence dQ = $\frac{5}{2}$ P_A(V_A-V_D)

This is a long answer type question as classified in NCERT Exemplar

pV^1/2= constant

P=k/ $\sqrt[]{V}$

Work done from 1 to 2

W= ${\int }_{V1}^{V2}pdV=K{\int }_{V1}^{V2}\frac{dV}{\sqrt[]{V}}=k{\left[\frac{\sqrt[]{V}}{\frac{1}{2}}\right]}_{V1}^{V2}=2K\left(\sqrt[]{V_2}-\sqrt[]{V_1}\right)$

from ideal equation = pV=nRT

T= pV/nR= $\frac{p\sqrt[]{V}\sqrt[]{V}}{nR}$

T= $\frac{K\sqrt[]{V}}{nR}$

T₁= $\frac{K\sqrt[]{V1}}{nR}$ , T₁= $\frac{K\sqrt[]{V2}}{nR}$

$\frac{T_1}{T_2}=\frac{\frac{k\sqrt[]{V_1}}{nR}}{\frac{k\sqrt[]{V_2}}{nR}}={\sqrt[]{\frac{V_1}{V_2}}}^{}$ = $\sqrt[]{\frac{V_1}{2V_1}}=\frac{1}{\sqrt[]{2}}$

U= $\frac{3}{2}RT$

$?U=U_2-U_1=\frac{3}{2}R\left(T_1-T_2\right)$

= $\frac{3}{2}$ RT₁( $\sqrt[]{V}-1$ )

$?W$ =2p₁V₁^1/2( $\sqrt[]{V_2}-\sqrt[]{V_2}$ )

 = 2p₁V₁^1/2(2 $\sqrt[]{V_1}-\sqrt[]{V_1}$ )

 = 2p₁V₁( $\sqrt[]{2}-1$ )= 2RT₁( $\sqrt[]{2}-1$ )

$?Q$ = $?U+?W$

= $\frac{3}{2}$ RT₁( $\sqrt[]{2}-1$ )+ 2RT₁( $\sqrt[]{2}-1$ )

= $\frac{7}{2}R{T}_1(\sqrt[]{2}-1)$