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PHXI12:THERMODYNAMICS

371297 One mole of a monoatomic ideal gas follows a process $A B$ , as shown. The specific heat of the process is $\frac{13 R}{6}$ . The value of $n$ on $P$ -axis is
supporting img

1 6

2 8

3 11

4 10

Explanation:

$C = \frac{13 R}{6} \Rightarrow C_{v} = \frac{3}{2} R$
$C = \frac{Δ Q}{Δ T} = \frac{(Δ U + Δ W)}{Δ T}$
$= \frac{Δ U}{Δ T} + \frac{Δ W}{Δ T} = C_{v} + \frac{Δ W}{Δ T}$
So, $\frac{Δ W}{Δ T} = C - C_{v} = \frac{13 R}{6} - \frac{3 R}{2} = \frac{2 R}{3}$
$Δ T = T_{2} - T_{1}$
$= \frac{(n P_{0}) (5 V_{0})}{R} - \frac{(3 P_{0}) V_{0}}{R} = \frac{P_{0} V_{0}}{R} (5 n - 3)$
$Δ W = \frac{(5 V_{0} - V_{0}) (n P_{0} + 3 P_{0})}{2} = 2 P_{0} V_{0} (n + 3)$
$\frac{Δ W}{Δ T} = \frac{2 (n + 3) R}{(5 n - 3)} = \frac{2 R}{3}$
On solving, $n = 6$

PHXI12:THERMODYNAMICS

371298 A gas at pressure $6 \times 10^{5} N m^{- 2}$ and volume $1 m^{3}$ and its pressure falls to $4 \times 10^{5} N m^{- 2}$ when its volume is $3 m^{3}$ . Given that the indicator diagram is a straight line, work done by the system is

1

6 \times 10^{5} J

2

3 \times 10^{5} J

3

4 \times 10^{5} J

4

10 \times 10^{5} J

Explanation:

supporting img

Work done by the system $=$ area of rectangle $B C D E +$ area of $Δ A B C$
$= 4 \times 10^{5} \times 2 + \frac{2 \times 10^{5} \times 2}{2}$
$W = 10 \times 10^{5} J$

PHXI12:THERMODYNAMICS

371299 An ideal gas undergoes four different processes from the same initial state as shown in the figure below. Those processes are adiabatic, isothermal, isobaric and isochoric. The curve which represents the adiabatic process among $1, 2, 3$ and 4 is
supporting img

1 2

2 3

3 4

4 1

Explanation:

1 : Isochoric $(∵ V =$ constant $)$
2: Adiabatic $(∵$ non linear $P V^{r} =$ constant $)$
3: Isothermal $(∵ P \propto \frac{1}{V} with T =$ constant )
4: Isobaric ( $∵ P =$ constant )

NEET - 2022

PHXI12:THERMODYNAMICS

371300 On a $T - P$ diagram, two moles of ideal gas perform process $A B$ and $C D$ . If the work done by the gas in the process $A B$ is two times the work done in the process $C D$ then what is the value of $T_{1} / T_{2}$ ?
supporting img

1 1

2

1 / 2

3 4

4 2

Explanation:

Let slope of line $A C$ is $m_{1} &$ that of line $D B$ is $m_{2}$ , then slope $(m) = \frac{T coordinate}{P coordinate}$
$\Rightarrow P$ -coordinate $= \frac{T - coordinate}{m}$ $P_{A} = \frac{T_{1}}{m_{1}}, P_{B} = \frac{T_{1}}{m_{2}}, P_{C} = \frac{T_{2}}{m_{1}} & P_{D} = \frac{T_{2}}{m_{2}}$
supporting img

Given that $W_{A B} = 2 W_{C D}$
$n R T_{1} \ln \frac{P_{A}}{P_{B}} = 2 n R T_{2} \ln \frac{P_{C}}{P_{D}}$
$n R T_{1} \ln \frac{m_{2}}{m_{1}} = 2 n R T_{2} \ln \frac{m_{2}}{m_{1}} \frac{T_{1}}{T_{2}} = 2$

PHXI12:THERMODYNAMICS

371297 One mole of a monoatomic ideal gas follows a process $A B$ , as shown. The specific heat of the process is $\frac{13 R}{6}$ . The value of $n$ on $P$ -axis is
supporting img

1 6

2 8

3 11

4 10

Explanation:

$C = \frac{13 R}{6} \Rightarrow C_{v} = \frac{3}{2} R$
$C = \frac{Δ Q}{Δ T} = \frac{(Δ U + Δ W)}{Δ T}$
$= \frac{Δ U}{Δ T} + \frac{Δ W}{Δ T} = C_{v} + \frac{Δ W}{Δ T}$
So, $\frac{Δ W}{Δ T} = C - C_{v} = \frac{13 R}{6} - \frac{3 R}{2} = \frac{2 R}{3}$
$Δ T = T_{2} - T_{1}$
$= \frac{(n P_{0}) (5 V_{0})}{R} - \frac{(3 P_{0}) V_{0}}{R} = \frac{P_{0} V_{0}}{R} (5 n - 3)$
$Δ W = \frac{(5 V_{0} - V_{0}) (n P_{0} + 3 P_{0})}{2} = 2 P_{0} V_{0} (n + 3)$
$\frac{Δ W}{Δ T} = \frac{2 (n + 3) R}{(5 n - 3)} = \frac{2 R}{3}$
On solving, $n = 6$

PHXI12:THERMODYNAMICS

371298 A gas at pressure $6 \times 10^{5} N m^{- 2}$ and volume $1 m^{3}$ and its pressure falls to $4 \times 10^{5} N m^{- 2}$ when its volume is $3 m^{3}$ . Given that the indicator diagram is a straight line, work done by the system is

1

6 \times 10^{5} J

2

3 \times 10^{5} J

3

4 \times 10^{5} J

4

10 \times 10^{5} J

Explanation:

supporting img

Work done by the system $=$ area of rectangle $B C D E +$ area of $Δ A B C$
$= 4 \times 10^{5} \times 2 + \frac{2 \times 10^{5} \times 2}{2}$
$W = 10 \times 10^{5} J$

PHXI12:THERMODYNAMICS

371299 An ideal gas undergoes four different processes from the same initial state as shown in the figure below. Those processes are adiabatic, isothermal, isobaric and isochoric. The curve which represents the adiabatic process among $1, 2, 3$ and 4 is
supporting img

1 2

2 3

3 4

4 1

Explanation:

1 : Isochoric $(∵ V =$ constant $)$
2: Adiabatic $(∵$ non linear $P V^{r} =$ constant $)$
3: Isothermal $(∵ P \propto \frac{1}{V} with T =$ constant )
4: Isobaric ( $∵ P =$ constant )

NEET - 2022

PHXI12:THERMODYNAMICS

371300 On a $T - P$ diagram, two moles of ideal gas perform process $A B$ and $C D$ . If the work done by the gas in the process $A B$ is two times the work done in the process $C D$ then what is the value of $T_{1} / T_{2}$ ?
supporting img

1 1

2

1 / 2

3 4

4 2

Explanation:

Let slope of line $A C$ is $m_{1} &$ that of line $D B$ is $m_{2}$ , then slope $(m) = \frac{T coordinate}{P coordinate}$
$\Rightarrow P$ -coordinate $= \frac{T - coordinate}{m}$ $P_{A} = \frac{T_{1}}{m_{1}}, P_{B} = \frac{T_{1}}{m_{2}}, P_{C} = \frac{T_{2}}{m_{1}} & P_{D} = \frac{T_{2}}{m_{2}}$
supporting img

Given that $W_{A B} = 2 W_{C D}$
$n R T_{1} \ln \frac{P_{A}}{P_{B}} = 2 n R T_{2} \ln \frac{P_{C}}{P_{D}}$
$n R T_{1} \ln \frac{m_{2}}{m_{1}} = 2 n R T_{2} \ln \frac{m_{2}}{m_{1}} \frac{T_{1}}{T_{2}} = 2$

PHXI12:THERMODYNAMICS

371297 One mole of a monoatomic ideal gas follows a process $A B$ , as shown. The specific heat of the process is $\frac{13 R}{6}$ . The value of $n$ on $P$ -axis is
supporting img

1 6

2 8

3 11

4 10

Explanation:

$C = \frac{13 R}{6} \Rightarrow C_{v} = \frac{3}{2} R$
$C = \frac{Δ Q}{Δ T} = \frac{(Δ U + Δ W)}{Δ T}$
$= \frac{Δ U}{Δ T} + \frac{Δ W}{Δ T} = C_{v} + \frac{Δ W}{Δ T}$
So, $\frac{Δ W}{Δ T} = C - C_{v} = \frac{13 R}{6} - \frac{3 R}{2} = \frac{2 R}{3}$
$Δ T = T_{2} - T_{1}$
$= \frac{(n P_{0}) (5 V_{0})}{R} - \frac{(3 P_{0}) V_{0}}{R} = \frac{P_{0} V_{0}}{R} (5 n - 3)$
$Δ W = \frac{(5 V_{0} - V_{0}) (n P_{0} + 3 P_{0})}{2} = 2 P_{0} V_{0} (n + 3)$
$\frac{Δ W}{Δ T} = \frac{2 (n + 3) R}{(5 n - 3)} = \frac{2 R}{3}$
On solving, $n = 6$

PHXI12:THERMODYNAMICS

371298 A gas at pressure $6 \times 10^{5} N m^{- 2}$ and volume $1 m^{3}$ and its pressure falls to $4 \times 10^{5} N m^{- 2}$ when its volume is $3 m^{3}$ . Given that the indicator diagram is a straight line, work done by the system is

1

6 \times 10^{5} J

2

3 \times 10^{5} J

3

4 \times 10^{5} J

4

10 \times 10^{5} J

Explanation:

supporting img

Work done by the system $=$ area of rectangle $B C D E +$ area of $Δ A B C$
$= 4 \times 10^{5} \times 2 + \frac{2 \times 10^{5} \times 2}{2}$
$W = 10 \times 10^{5} J$

PHXI12:THERMODYNAMICS

371299 An ideal gas undergoes four different processes from the same initial state as shown in the figure below. Those processes are adiabatic, isothermal, isobaric and isochoric. The curve which represents the adiabatic process among $1, 2, 3$ and 4 is
supporting img

1 2

2 3

3 4

4 1

Explanation:

1 : Isochoric $(∵ V =$ constant $)$
2: Adiabatic $(∵$ non linear $P V^{r} =$ constant $)$
3: Isothermal $(∵ P \propto \frac{1}{V} with T =$ constant )
4: Isobaric ( $∵ P =$ constant )

NEET - 2022

PHXI12:THERMODYNAMICS

371300 On a $T - P$ diagram, two moles of ideal gas perform process $A B$ and $C D$ . If the work done by the gas in the process $A B$ is two times the work done in the process $C D$ then what is the value of $T_{1} / T_{2}$ ?
supporting img

1 1

2

1 / 2

3 4

4 2

Explanation:

Let slope of line $A C$ is $m_{1} &$ that of line $D B$ is $m_{2}$ , then slope $(m) = \frac{T coordinate}{P coordinate}$
$\Rightarrow P$ -coordinate $= \frac{T - coordinate}{m}$ $P_{A} = \frac{T_{1}}{m_{1}}, P_{B} = \frac{T_{1}}{m_{2}}, P_{C} = \frac{T_{2}}{m_{1}} & P_{D} = \frac{T_{2}}{m_{2}}$
supporting img

Given that $W_{A B} = 2 W_{C D}$
$n R T_{1} \ln \frac{P_{A}}{P_{B}} = 2 n R T_{2} \ln \frac{P_{C}}{P_{D}}$
$n R T_{1} \ln \frac{m_{2}}{m_{1}} = 2 n R T_{2} \ln \frac{m_{2}}{m_{1}} \frac{T_{1}}{T_{2}} = 2$

PHXI12:THERMODYNAMICS

371297 One mole of a monoatomic ideal gas follows a process $A B$ , as shown. The specific heat of the process is $\frac{13 R}{6}$ . The value of $n$ on $P$ -axis is
supporting img

1 6

2 8

3 11

4 10

Explanation:

$C = \frac{13 R}{6} \Rightarrow C_{v} = \frac{3}{2} R$
$C = \frac{Δ Q}{Δ T} = \frac{(Δ U + Δ W)}{Δ T}$
$= \frac{Δ U}{Δ T} + \frac{Δ W}{Δ T} = C_{v} + \frac{Δ W}{Δ T}$
So, $\frac{Δ W}{Δ T} = C - C_{v} = \frac{13 R}{6} - \frac{3 R}{2} = \frac{2 R}{3}$
$Δ T = T_{2} - T_{1}$
$= \frac{(n P_{0}) (5 V_{0})}{R} - \frac{(3 P_{0}) V_{0}}{R} = \frac{P_{0} V_{0}}{R} (5 n - 3)$
$Δ W = \frac{(5 V_{0} - V_{0}) (n P_{0} + 3 P_{0})}{2} = 2 P_{0} V_{0} (n + 3)$
$\frac{Δ W}{Δ T} = \frac{2 (n + 3) R}{(5 n - 3)} = \frac{2 R}{3}$
On solving, $n = 6$

PHXI12:THERMODYNAMICS

371298 A gas at pressure $6 \times 10^{5} N m^{- 2}$ and volume $1 m^{3}$ and its pressure falls to $4 \times 10^{5} N m^{- 2}$ when its volume is $3 m^{3}$ . Given that the indicator diagram is a straight line, work done by the system is

1

6 \times 10^{5} J

2

3 \times 10^{5} J

3

4 \times 10^{5} J

4

10 \times 10^{5} J

Explanation:

supporting img

Work done by the system $=$ area of rectangle $B C D E +$ area of $Δ A B C$
$= 4 \times 10^{5} \times 2 + \frac{2 \times 10^{5} \times 2}{2}$
$W = 10 \times 10^{5} J$

PHXI12:THERMODYNAMICS

371299 An ideal gas undergoes four different processes from the same initial state as shown in the figure below. Those processes are adiabatic, isothermal, isobaric and isochoric. The curve which represents the adiabatic process among $1, 2, 3$ and 4 is
supporting img

1 2

2 3

3 4

4 1

Explanation:

1 : Isochoric $(∵ V =$ constant $)$
2: Adiabatic $(∵$ non linear $P V^{r} =$ constant $)$
3: Isothermal $(∵ P \propto \frac{1}{V} with T =$ constant )
4: Isobaric ( $∵ P =$ constant )

NEET - 2022

PHXI12:THERMODYNAMICS

371300 On a $T - P$ diagram, two moles of ideal gas perform process $A B$ and $C D$ . If the work done by the gas in the process $A B$ is two times the work done in the process $C D$ then what is the value of $T_{1} / T_{2}$ ?
supporting img

1 1

2

1 / 2

3 4

4 2

Explanation:

Let slope of line $A C$ is $m_{1} &$ that of line $D B$ is $m_{2}$ , then slope $(m) = \frac{T coordinate}{P coordinate}$
$\Rightarrow P$ -coordinate $= \frac{T - coordinate}{m}$ $P_{A} = \frac{T_{1}}{m_{1}}, P_{B} = \frac{T_{1}}{m_{2}}, P_{C} = \frac{T_{2}}{m_{1}} & P_{D} = \frac{T_{2}}{m_{2}}$
supporting img

Given that $W_{A B} = 2 W_{C D}$
$n R T_{1} \ln \frac{P_{A}}{P_{B}} = 2 n R T_{2} \ln \frac{P_{C}}{P_{D}}$
$n R T_{1} \ln \frac{m_{2}}{m_{1}} = 2 n R T_{2} \ln \frac{m_{2}}{m_{1}} \frac{T_{1}}{T_{2}} = 2$

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