## A quantity of an ideal monoatomic gas consists of n moles initially at temperature T1 …..

Q: COMPREHENSION:
A quantity of an ideal monoatomic gas consists of n moles initially at temperature T1 . The pressure and volume are then slowly doubled in such a manner so as to trace out a straight line on a P-V diagram

1 . For this process, the ratio $\frac{W}{nRT_1}$ is equal to (where W is work done by the gas) :
(A) 1.5
(B) 3
(C) 4.5
(D) 6
2. For the same process, the ratio $\frac{Q}{nRT_1}$ is equal to (where Q is heat supplied to the gas) :
(A) 1.5
(B) 3
(C) 4.5
(D) 6

3. If C is defined as the average molar specific heat for the process then $\frac{C}{R}$ has value
(A) 1.5
(B) 2
(C) 3
(D) 6

Ans: 1:(A) ; 2: (D) ; 3: (B)

Solution: 1: W = Area under the curve $= \frac{3}{2}P_1 V_1$

$P_1 V_1 = n R T_1$

Therefore $\frac{W}{n R T_1} = \frac{\frac{3}{2}P_1 V_1}{P_1 V_1}$

2: Q = dU + W
dU = n Cv dT
For final state P2V2 = 2P1 2V1 = 4P1V1 = nR(4T1)
Hence final temp. is 4T1

$dU = n \frac{3}{2}R.3T_1 = \frac{9}{2}n R T_1$

$Q = \frac{3}{2}n R T_1 + \frac{9}{2}n R T_1 = 6 n R T_1$

$\frac{Q}{n R T_1} = 6$

3: n C ΔT = Q

⇒ n C ΔT = 6 n R T1

dT = 4T1 – T1 = 3T1

n . C . 3 T1 = 6 n R T1

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

## A gas undergoes an adiabatic process and an isothermal process. The two processes are plotted on a P-V diagram….

Q: A gas undergoes an adiabatic process and an isothermal process. The two processes are plotted on a P-V diagram. The resulting curves intersect at a point P. Tangents are drawn to the two curves at P. These make angles of 135º & 121º with the positive V-axis. If tan 59º = 5/3, the gas is likely

(A) monoatomic

(B) diatomic

(C) triatomic

(D) a mixture of monoatomic & diatomic gases

Ans: (A)

Solution: The slope of isothermal curve at point of intersection is

$\displaystyle \frac{dP}{dV} = -\frac{P}{V} = tan135^o$ …(i)

The slope of adiabatic curve at point of intersection is

$\displaystyle \frac{dP}{dV} = -\gamma \frac{P}{V} = tan121^o$ …(ii)

From (i) & (ii)

from (1) and (2)

γ = tan 59° = 1.66 = 5/3

Hence , gas is monoatomic

## An ideal gas undergoes a cyclic process abcda which is shown by pressure density curve….

Q: An ideal gas undergoes a cyclic process abcda which is shown by pressure density curve.

(A) Work done by the gas in the process ‘bc’ is zero

(B) Work done by the gas in the process ‘cd’ is negative

(C) Internal energy of the gas at point ‘a’ is greater than at state ‘c’

(D) Net work done by the gas in the cycle is negative.

Ans: (A),(B)

Solution: $\frac{P}{\rho} = \frac{R}{M_0}T$

Slope of the curve ∝ Temperature

Hence cd and ab are isothermal processes.

i.e. bc and da are constant volume process
(A) and (B) are true. Temp. in cd process is greater than ab. Net work done by the gas in the cycle is
negative, as is clear by the PV-diagram.

$\frac{P}{\rho} = \frac{R}{M_0}T$

## In the figure shown the pressure of the gas in state B is

Q: In the figure shown the pressure of the gas in state B is:

(a) $\frac{63}{25}P_0$

(b) $\frac{73}{25}P_0$

(c) $\frac{48}{25}P_0$

(d) None of these

Ans: (b)

Solution: $AN = 3V_0 cos37^o$

$P_B = \frac{P_0}{V_0}(V_0 + 3 V_0 \times \frac{16}{25})$

$= P_0 (1 + \frac{48}{25}) = (\frac{73}{25})P_0$

## An Ideal gas undergoes a four step cycle as shown in P-V diagram below . During this cycle …

Q: An Ideal gas undergoes a four step cycle as shown in P-V diagram below . During this cycle , heat is absorbed by the gas in

(a) Steps 1 and 2

(b) Steps 1 and 3

(c) Steps 1 and 4

(d) Steps 2 and 4

Ans: (c)

Solution:Process 1:

P = constant , Volume increases and temperature also increases

Q = W + ΔU

W = +ve , ΔU = positive

Heat is positive and supplied to the gas

Process 2 :
V = constant , Pressure decreases

T ∝ P , (V = constant )

Temperature decreases , W = 0
ΔT is negative and ΔU = f/2 nRΔT

ΔU is also negative

Q = ΔU + W

Heat is negative and rejected by gas

Process 3 :
P = constant , Volume decreases

Temperature also decreases

W = PΔV = negative

ΔU = f/2 (nRΔT) = negative

Heat is negative and rejected by gas

Process 4 :
V = constant , Pressure increases

W = 0
PV = n RT => temperature increases

ΔU = f/2 (nRΔT) is positive

ΔQ = ΔU + W = Positive

## Consider one mole of helium gas enclosed in a container at initial pressure P1 and volume V1 . It expands Isothermally …..

Q: Consider one mole of helium gas enclosed in a container at initial pressure P1 and volume V1 . It expands Isothermally to volume 4V1 . After this gas expands adiabatically and its volume becomes 32 V1 . The work done by the gas during isothermal and adiabatic expansion processes are Wiso and Wadia respectively . If the ratio $\frac{W_{iso}}{W_{adia}} = f ln2$ , then f is —–

Ans: (1.77 )

Solution: $\frac{P_1}{4}(4V_1)^{5/3} = P_2 (32 V_1)^{5/3}$

$P_2 = \frac{P_1}{4}(\frac{1}{8})^{5/3} = \frac{P_1}{128}$

$\displaystyle W_{adia} = \frac{P_1 V_1 – P_2 V_2}{\gamma -1}$

$\displaystyle W_{adia} = \frac{P_1 V_1 – (P_1/128) (32 V_1)}{\frac{5}{3} -1}$

$\displaystyle W_{adia} = \frac{P_1 V_1 (3/4)}{\frac{2}{3}}$

$\displaystyle W_{adia} = \frac{9}{8}P_1 V_1$

$\displaystyle W_{iso} = P_1 V_1 ln(\frac{4V_1}{V_1})$

$\displaystyle W_{iso} = 2 P_1 V_1 ln2$

$\displaystyle \frac{W_{iso}}{W_{adia}} = \frac{16}{9} ln 2$

f = 16/9 = 1.77

## If minimum possible work is done by a refrigerator in converting 100 gm of water at 0°C to ice ….

Q: If minimum possible work is done by a refrigerator in converting 100 gm of water at 0°C to ice , how much heat (in calorie) is released to the surroundings at temperature 27 °C (Latent heat of ice = 80 cal/gram) to the nearest integer ?

Click to See Solution :
Ans: (8791)

Sol: T1 = 27 °C = 300 K , Lice = 80 cal/g

Heat energy extracted from water , Q2 = m L

Q2 = 100 x 80 = 8000 cal

T2 = 0 °C = 273 K

$\displaystyle \frac{Q_1}{Q_2} = \frac{T_1}{T_2}$

$\displaystyle Q_1 = Q_2 \times \frac{T_1}{T_2}$

$\displaystyle Q_1 = 8000 \times \frac{300}{273}$

= 8791 cal

## Two ideal Carnot engines operate in cascade (all heat given up by one engine is used by the other engine to produce work ) …

Q: Two ideal Carnot engines operate in cascade (all heat given up by one engine is used by the other engine to produce work ) between temperatures T1 and T2 . The temperature of the hot reservoir of the first engine is T1 and temperature of cold reservoir of the second engine is T2 . T is the temperature of sink of first engine which is also the source for the second engine . How is T related to T1 and T2 , if both the engines perform equal amount to work ?

(a) $\displaystyle T = \sqrt{T_1 T_2}$

(b) T = 0

(c) $\displaystyle T = \frac{2T_1 T_2}{T_1 + T_2}$

$\displaystyle T = \frac{T_1+ T_2}{2}$

Click to See Solution :
Ans: (d)
Sol: For first engine ,

$\displaystyle W_1 = Q_1 – Q_2$

and $\displaystyle \frac{Q_2}{Q_1} = \frac{T}{T_1}$

$\displaystyle W_1 = Q_1[1 – \frac{Q_2}{Q_1} ]$

$\displaystyle W_1 = Q_1[1 – \frac{T}{T_1} ]$ ..(i)

For Second engine ,

$\displaystyle W’ = Q_2 – Q_2′$

and $\displaystyle \frac{Q_2′}{Q_2} = \frac{T_2}{T}$

$\displaystyle W_2 = Q_2 [1-\frac{Q_2′}{Q_2}]$

$\displaystyle W_2 = Q_2 [1-\frac{T_2}{T}]$ …(ii)

Work done by first engine(W_1) = work done by second engine (W_2)

$\displaystyle Q_1[1 – \frac{T}{T_1} ] = Q_2 [1-\frac{T_2}{T}]$

$\displaystyle \frac{Q_1}{Q_2} = \frac{1-\frac{T_2}{T}}{1-\frac{T}{T_1}}$

$\displaystyle \frac{T_1}{T} = \frac{1-\frac{T_2}{T}}{1-\frac{T}{T_1}}$

$\displaystyle T_1 – T = T – T_2$

$\displaystyle T = \frac{T_1 + T_2}{2}$

## Consider a heat engine working with a hot reservoir of the furnace gases at 2100 °C, when the cooling water is available at -35.7 °C. then, choose the correct option(s).

Q: Consider a heat engine working with a hot reservoir of the furnace gases at 2100 °C , when the cooling water is available at -35.7 °C. then, choose the correct option(s).

(a) Engine efficiency can be made more than 90% without altering operating temperatures

(b) Engine efficiency always remains less than 90 %

(c) Engine efficiency is much lower practically due to irreversibility in the actual cycle

(d) Engine efficiency is 90% but such temperatures are not available practically

Ans: (c)

Sol: Temperature of hot Reservoir , T1 = 2100 + 273 = 2373 k

Temperature of cold Reservoir , T2 = -35.7 + 273 = 237.3 K

Maximum Efficiency , $\displaystyle \eta_{max} = 1 – \frac{T_2}{T_1}$

$\displaystyle \eta_{max} = 1 – \frac{237.3}{2373}$

$\displaystyle \eta_{max} = 1 – \frac{1}{10} = \frac{9}{10}$

Percentage efficiency $\displaystyle = \frac{9}{10} \times 100$

= 90 %

## During a process a system receives 30 kJ of heat from a reservoir and does 60 kJ of work. Then, choose the correct option(s).

Q: During a process a system receives 30 kJ of heat from a reservoir and does 60 kJ of work. Then, choose the correct option(s).

(a) It is possible to reach initial state by a quasistatic isothermal process

(b) It is possible to reach initial state using a constant volume process

(c) It is possible to reach initial state using a constant pressure process

(d) It is possible to reach initial state using an adiabatic process

Ans: (d)
Sol: Here ΔQ = 30 kJ , ΔW = 60 kJ

For Process 1-2 ,

ΔQ1-2 = ΔU1-2 + ΔW1-2

30 = ΔU1-2 + 60

ΔU1-2 = – 30 k J

For Process 2-1 ,

ΔQ2-1 = ΔU2-1 + ΔW2-1

0 = U1 – U2 + ΔW2-1 (for Adiabatic Process , ΔQ = 0 )

0 = 30 + ΔW2-1

ΔW2-1 = – 30 kJ

Therefore , by doing a work of 30 kJ on the System , it can go back to Initial state