Thermodynamics

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For adiabatic change process we know

P₁V₁^y= P₂V₂^y

P (V+ $?V$ )^y = (P+ $?P$ )V^y

PV^y (1+ $\frac{?V}{V}$ ) ^y=p (1+ $\frac{?P}{P}$ )V^y

PV^y (1+ $\frac{?V}{V}$ ) $\approx$ PV^y (1+ $\frac{?V}{V}$ )

Y $\frac{?V}{V}=\frac{?P}{P}$

dV= $\frac{1}{Y}\frac{V}{p}dP$

hence work done increasing the pressure from P₁ to P₂

W= ${\int }_{p1}^{p2}pdV={\int }_{p2}^{p1}p\frac{1}{Y}\frac{V}{p}dp$

= $\frac{V}{Y}\left(P_2-P_1\right)$

W= $\frac{P_2-P_1}{Y}V$

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

Height of stairs h= 10m

Energy produced by burning 1 kg of fat = 7000Kcal

Energy produced by burning 5kg of fat = 5 $*7000=35000Kcal$

Energy utilised in going up and down one time

= mgh + $\frac{1}{2}mgh$ = $\frac{3}{2}mgh$

= $\frac{3}{2}*60*10*10$

= 9000J= 9000/4.2=3000/1.4cal

Number of times, the person has to go up and down the stairs

= $\frac{35*{10}^6}{\frac{3000}{1.4}}=\frac{3.5*1.4*{10}^6}{3000}$  = 16.3 $*{10}^3$ times

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Temperature of the source T₁= 500K and sink T₂= 300K

Work done W= 1000J

Efficiency of Carnot engine = 1-T₂/T₁= 1-300/500= 200/500= 2/5

Efficiency = W/Q₁

So Q₁= W/efficiency = 1000 $\left(\frac{5}{2}\right)=2500J\mathit{}$

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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.

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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.

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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

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

So its working against the surroundings.

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(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

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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 $\frac{P_o}{V_o}$ )