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Thermodynamics

148508 During an experiment, an ideal gas is found to obey an additional law $V^{2} =$ constant. The gas is initially at temperature $T$ and volume. $V$ . The temperature of the gas will be following, when it expands to a volume $2 V$ ?

1

\sqrt{2} T

2

\sqrt{4} T

3

\sqrt{6} T

4

\sqrt{5} T

Explanation:

A Given,
$\begin{array}{ll} We know {VP}^{2} = constant \\ Again PV = nRT \\ P = \frac{nRT}{V} \end{array}$
Put in equation (i), We get,
$∴ \frac{(nRT)^{2}}{V} = constant$
$\frac{T^{2}}{V} = constant$
Thus volume $V$ when expanded to $2 V$ , temperature $T_{2}$
$\frac{T_{2}^{2}}{T_{1}^{2}} = \frac{V_{2}}{V_{1}} = \frac{2 V}{V}$
$T_{2} = \sqrt{2} T_{1}$

CG PET- 2014

Thermodynamics

148509 Two containers of equal volume contain the same gas at pressure $P_{1}$ and $P_{2}$ and absolute temperature $T_{1}$ and $T_{2}$ respectively. On joining the vessels the gas reaches a common pressure $P$ and common temperature $T$ . The ratio $P / T$ is equal to

1

\frac{P_{1}}{T_{1}} + \frac{P_{2}}{T_{2}}

2

\frac{P_{1} T_{1} + P_{2} T_{2}}{{(T_{1} + T_{2})}^{2}}

3

\frac{P_{1} T_{2} + P_{2} T_{1}}{{(T_{1} + T_{2})}^{2}}

4

\frac{P_{1}}{2 T_{1}} + \frac{P_{2}}{2 T_{2}}

Explanation:

D From ideal gas-
$PV = mRT$
$∴ m = \frac{PV}{RT}$
$∴$ Number of moles of gas in first container-
$m_{1} = \frac{P_{1} V}{{RT}_{1}}$
Number of moles of gas in second container-
$m_{2} = \frac{P_{2} V}{{RT}_{2}}$
Number of moles in containers when joined with each other.
$m = \frac{PV}{RT}$
But,
$m = m_{1} + m_{2}$
$\frac{P (2 V)}{RT} = \frac{P_{1} V}{{RT}_{1}} + \frac{P_{2} V}{{RT}_{2}}$
$\frac{2 P}{T} = \frac{P_{1}}{T_{1}} + \frac{P_{2}}{T_{2}}$
$\frac{P}{T} = \frac{P_{1}}{2 T_{1}} + \frac{P_{2}}{2 T_{2}}$

CG PET- 2006

Thermodynamics

148510 2 moles of an ideal mono-atomic gas is carried from a state $(P_{0}, V_{0})$ to state $(2 P_{0}, 2 V_{0})$ along a straight line path in a $P - V$ diagram. The amount of heat absorbed by the gas in the process is given by

1

3 P_{0} V_{0}

2

\frac{9}{2} P_{0} V_{0}

3

6 P_{0} V_{0}

4

\frac{3}{2} P_{0} V_{0}

Explanation:

C
Area under the curve given work done in $P - v$ graph
$W = V_{0} P_{0} + \frac{V_{0} P_{0}}{2}$
$W = \frac{3}{2} P_{0} V_{0}$
$Δ u = {nC}_{v} (T_{B} - T_{A})$
$= \frac{3 nR}{2} (\frac{2 P_{0} \times 2 V_{0}}{nR} - \frac{P_{0} V_{0}}{nR})$
$= \frac{9}{2} P_{0} V_{0}$
$Q = U + W$
$[according to first law of thermodynamic ]$
$Q = \frac{9}{2} P_{0} V_{0} + \frac{3}{2} P_{0} V_{0}$
$Q = 6 P_{0} V_{0}$

WB JEE 2017

Thermodynamics

148512 If the coefficient of superficial expansion is $x$ times the coefficient of cubical expansion, then the value of $x$ is

1 2

2

\frac{3}{2}

3

\frac{2}{3}

4

\frac{1}{2}

Explanation:

B Given that,
Coefficient of superficial expansion $= γ$
As we know that coefficient of cubical expansion is related to coefficient of superficial expansion.
The relation between cubical expansion and superficial expansion $β$ is given as
$\frac{β}{2} = \frac{γ}{3}$
$or γ = \frac{3 β}{2}$
It is given that in question coefficient of the cubical expansion is increased by the $x$ times of the coefficient of superficial expansion i.e.
$γ = β x$
From equation (i) and (ii),
$x = \frac{3}{2}$
$x = 1.5$

SCRA-2015

Thermodynamics

148513 ' $m$ ' grams of a gas of molecular weight $M$ is flowing in an isolated tube velocity $v$ . If the gas flow is suddenly stopped the rise in the temperature is : $(γ =$ ratio of specific heats, $R =$ universal gas constant, $J =$ mechanical equivalent of heat).

1

\frac{{Mv}^{2} (γ - 1)}{2 RJ}

2

\frac{m}{M} \frac{v^{2} (γ - 1)}{2 RJ}

3

\frac{m v^{2} γ}{2 RJ}

4

\frac{{Mv}^{2} γ}{2 RJ}

Explanation:

A Given that,
Mass of gas $= m$
Molecular weight $= M$
Velocity $= v$
Mechanical equivalent of heat $= J$
Since the gas flow is suddenly stopped, we will consider it to be an adiabatic process.
Change in energy $Δ$ K.E. $= \frac{1}{2} {mv}^{2} - \frac{1}{2} M .0$
$= \frac{1}{2} {mv}^{2}$
Work done for adiabatic process,
$W = \frac{nR Δ T}{γ - 1} W = \frac{P_{1} V_{1} - P_{2} V_{2}}{γ - 1}$
$Δ U = Δ W [for adiabatic process]$
$\frac{1}{2} {mv}^{2} = J \frac{nR Δ T}{γ - 1}$
$n = J \frac{m}{M} [$ no.of moles. $]$
$\frac{1}{2} {mv}^{2} = J \frac{mR Δ T}{M (γ - 1)}$
$\frac{v^{2}}{2} = \frac{JR Δ T}{M (γ - 1)}$
$Δ T = \frac{{Mv}^{2} (γ - 1)}{2 RJ}$

AP EAMCET(Medical)-2006

Thermodynamics

148508 During an experiment, an ideal gas is found to obey an additional law $V^{2} =$ constant. The gas is initially at temperature $T$ and volume. $V$ . The temperature of the gas will be following, when it expands to a volume $2 V$ ?

1

\sqrt{2} T

2

\sqrt{4} T

3

\sqrt{6} T

4

\sqrt{5} T

Explanation:

A Given,
$\begin{array}{ll} We know {VP}^{2} = constant \\ Again PV = nRT \\ P = \frac{nRT}{V} \end{array}$
Put in equation (i), We get,
$∴ \frac{(nRT)^{2}}{V} = constant$
$\frac{T^{2}}{V} = constant$
Thus volume $V$ when expanded to $2 V$ , temperature $T_{2}$
$\frac{T_{2}^{2}}{T_{1}^{2}} = \frac{V_{2}}{V_{1}} = \frac{2 V}{V}$
$T_{2} = \sqrt{2} T_{1}$

CG PET- 2014

Thermodynamics

148509 Two containers of equal volume contain the same gas at pressure $P_{1}$ and $P_{2}$ and absolute temperature $T_{1}$ and $T_{2}$ respectively. On joining the vessels the gas reaches a common pressure $P$ and common temperature $T$ . The ratio $P / T$ is equal to

1

\frac{P_{1}}{T_{1}} + \frac{P_{2}}{T_{2}}

2

\frac{P_{1} T_{1} + P_{2} T_{2}}{{(T_{1} + T_{2})}^{2}}

3

\frac{P_{1} T_{2} + P_{2} T_{1}}{{(T_{1} + T_{2})}^{2}}

4

\frac{P_{1}}{2 T_{1}} + \frac{P_{2}}{2 T_{2}}

Explanation:

D From ideal gas-
$PV = mRT$
$∴ m = \frac{PV}{RT}$
$∴$ Number of moles of gas in first container-
$m_{1} = \frac{P_{1} V}{{RT}_{1}}$
Number of moles of gas in second container-
$m_{2} = \frac{P_{2} V}{{RT}_{2}}$
Number of moles in containers when joined with each other.
$m = \frac{PV}{RT}$
But,
$m = m_{1} + m_{2}$
$\frac{P (2 V)}{RT} = \frac{P_{1} V}{{RT}_{1}} + \frac{P_{2} V}{{RT}_{2}}$
$\frac{2 P}{T} = \frac{P_{1}}{T_{1}} + \frac{P_{2}}{T_{2}}$
$\frac{P}{T} = \frac{P_{1}}{2 T_{1}} + \frac{P_{2}}{2 T_{2}}$

CG PET- 2006

Thermodynamics

148510 2 moles of an ideal mono-atomic gas is carried from a state $(P_{0}, V_{0})$ to state $(2 P_{0}, 2 V_{0})$ along a straight line path in a $P - V$ diagram. The amount of heat absorbed by the gas in the process is given by

1

3 P_{0} V_{0}

2

\frac{9}{2} P_{0} V_{0}

3

6 P_{0} V_{0}

4

\frac{3}{2} P_{0} V_{0}

Explanation:

C
Area under the curve given work done in $P - v$ graph
$W = V_{0} P_{0} + \frac{V_{0} P_{0}}{2}$
$W = \frac{3}{2} P_{0} V_{0}$
$Δ u = {nC}_{v} (T_{B} - T_{A})$
$= \frac{3 nR}{2} (\frac{2 P_{0} \times 2 V_{0}}{nR} - \frac{P_{0} V_{0}}{nR})$
$= \frac{9}{2} P_{0} V_{0}$
$Q = U + W$
$[according to first law of thermodynamic ]$
$Q = \frac{9}{2} P_{0} V_{0} + \frac{3}{2} P_{0} V_{0}$
$Q = 6 P_{0} V_{0}$

WB JEE 2017

Thermodynamics

148512 If the coefficient of superficial expansion is $x$ times the coefficient of cubical expansion, then the value of $x$ is

1 2

2

\frac{3}{2}

3

\frac{2}{3}

4

\frac{1}{2}

Explanation:

B Given that,
Coefficient of superficial expansion $= γ$
As we know that coefficient of cubical expansion is related to coefficient of superficial expansion.
The relation between cubical expansion and superficial expansion $β$ is given as
$\frac{β}{2} = \frac{γ}{3}$
$or γ = \frac{3 β}{2}$
It is given that in question coefficient of the cubical expansion is increased by the $x$ times of the coefficient of superficial expansion i.e.
$γ = β x$
From equation (i) and (ii),
$x = \frac{3}{2}$
$x = 1.5$

SCRA-2015

Thermodynamics

148513 ' $m$ ' grams of a gas of molecular weight $M$ is flowing in an isolated tube velocity $v$ . If the gas flow is suddenly stopped the rise in the temperature is : $(γ =$ ratio of specific heats, $R =$ universal gas constant, $J =$ mechanical equivalent of heat).

1

\frac{{Mv}^{2} (γ - 1)}{2 RJ}

2

\frac{m}{M} \frac{v^{2} (γ - 1)}{2 RJ}

3

\frac{m v^{2} γ}{2 RJ}

4

\frac{{Mv}^{2} γ}{2 RJ}

Explanation:

A Given that,
Mass of gas $= m$
Molecular weight $= M$
Velocity $= v$
Mechanical equivalent of heat $= J$
Since the gas flow is suddenly stopped, we will consider it to be an adiabatic process.
Change in energy $Δ$ K.E. $= \frac{1}{2} {mv}^{2} - \frac{1}{2} M .0$
$= \frac{1}{2} {mv}^{2}$
Work done for adiabatic process,
$W = \frac{nR Δ T}{γ - 1} W = \frac{P_{1} V_{1} - P_{2} V_{2}}{γ - 1}$
$Δ U = Δ W [for adiabatic process]$
$\frac{1}{2} {mv}^{2} = J \frac{nR Δ T}{γ - 1}$
$n = J \frac{m}{M} [$ no.of moles. $]$
$\frac{1}{2} {mv}^{2} = J \frac{mR Δ T}{M (γ - 1)}$
$\frac{v^{2}}{2} = \frac{JR Δ T}{M (γ - 1)}$
$Δ T = \frac{{Mv}^{2} (γ - 1)}{2 RJ}$

AP EAMCET(Medical)-2006

Thermodynamics

148508 During an experiment, an ideal gas is found to obey an additional law $V^{2} =$ constant. The gas is initially at temperature $T$ and volume. $V$ . The temperature of the gas will be following, when it expands to a volume $2 V$ ?

1

\sqrt{2} T

2

\sqrt{4} T

3

\sqrt{6} T

4

\sqrt{5} T

Explanation:

A Given,
$\begin{array}{ll} We know {VP}^{2} = constant \\ Again PV = nRT \\ P = \frac{nRT}{V} \end{array}$
Put in equation (i), We get,
$∴ \frac{(nRT)^{2}}{V} = constant$
$\frac{T^{2}}{V} = constant$
Thus volume $V$ when expanded to $2 V$ , temperature $T_{2}$
$\frac{T_{2}^{2}}{T_{1}^{2}} = \frac{V_{2}}{V_{1}} = \frac{2 V}{V}$
$T_{2} = \sqrt{2} T_{1}$

CG PET- 2014

Thermodynamics

148509 Two containers of equal volume contain the same gas at pressure $P_{1}$ and $P_{2}$ and absolute temperature $T_{1}$ and $T_{2}$ respectively. On joining the vessels the gas reaches a common pressure $P$ and common temperature $T$ . The ratio $P / T$ is equal to

1

\frac{P_{1}}{T_{1}} + \frac{P_{2}}{T_{2}}

2

\frac{P_{1} T_{1} + P_{2} T_{2}}{{(T_{1} + T_{2})}^{2}}

3

\frac{P_{1} T_{2} + P_{2} T_{1}}{{(T_{1} + T_{2})}^{2}}

4

\frac{P_{1}}{2 T_{1}} + \frac{P_{2}}{2 T_{2}}

Explanation:

D From ideal gas-
$PV = mRT$
$∴ m = \frac{PV}{RT}$
$∴$ Number of moles of gas in first container-
$m_{1} = \frac{P_{1} V}{{RT}_{1}}$
Number of moles of gas in second container-
$m_{2} = \frac{P_{2} V}{{RT}_{2}}$
Number of moles in containers when joined with each other.
$m = \frac{PV}{RT}$
But,
$m = m_{1} + m_{2}$
$\frac{P (2 V)}{RT} = \frac{P_{1} V}{{RT}_{1}} + \frac{P_{2} V}{{RT}_{2}}$
$\frac{2 P}{T} = \frac{P_{1}}{T_{1}} + \frac{P_{2}}{T_{2}}$
$\frac{P}{T} = \frac{P_{1}}{2 T_{1}} + \frac{P_{2}}{2 T_{2}}$

CG PET- 2006

Thermodynamics

148510 2 moles of an ideal mono-atomic gas is carried from a state $(P_{0}, V_{0})$ to state $(2 P_{0}, 2 V_{0})$ along a straight line path in a $P - V$ diagram. The amount of heat absorbed by the gas in the process is given by

1

3 P_{0} V_{0}

2

\frac{9}{2} P_{0} V_{0}

3

6 P_{0} V_{0}

4

\frac{3}{2} P_{0} V_{0}

Explanation:

C
Area under the curve given work done in $P - v$ graph
$W = V_{0} P_{0} + \frac{V_{0} P_{0}}{2}$
$W = \frac{3}{2} P_{0} V_{0}$
$Δ u = {nC}_{v} (T_{B} - T_{A})$
$= \frac{3 nR}{2} (\frac{2 P_{0} \times 2 V_{0}}{nR} - \frac{P_{0} V_{0}}{nR})$
$= \frac{9}{2} P_{0} V_{0}$
$Q = U + W$
$[according to first law of thermodynamic ]$
$Q = \frac{9}{2} P_{0} V_{0} + \frac{3}{2} P_{0} V_{0}$
$Q = 6 P_{0} V_{0}$

WB JEE 2017

Thermodynamics

148512 If the coefficient of superficial expansion is $x$ times the coefficient of cubical expansion, then the value of $x$ is

1 2

2

\frac{3}{2}

3

\frac{2}{3}

4

\frac{1}{2}

Explanation:

B Given that,
Coefficient of superficial expansion $= γ$
As we know that coefficient of cubical expansion is related to coefficient of superficial expansion.
The relation between cubical expansion and superficial expansion $β$ is given as
$\frac{β}{2} = \frac{γ}{3}$
$or γ = \frac{3 β}{2}$
It is given that in question coefficient of the cubical expansion is increased by the $x$ times of the coefficient of superficial expansion i.e.
$γ = β x$
From equation (i) and (ii),
$x = \frac{3}{2}$
$x = 1.5$

SCRA-2015

Thermodynamics

148513 ' $m$ ' grams of a gas of molecular weight $M$ is flowing in an isolated tube velocity $v$ . If the gas flow is suddenly stopped the rise in the temperature is : $(γ =$ ratio of specific heats, $R =$ universal gas constant, $J =$ mechanical equivalent of heat).

1

\frac{{Mv}^{2} (γ - 1)}{2 RJ}

2

\frac{m}{M} \frac{v^{2} (γ - 1)}{2 RJ}

3

\frac{m v^{2} γ}{2 RJ}

4

\frac{{Mv}^{2} γ}{2 RJ}

Explanation:

A Given that,
Mass of gas $= m$
Molecular weight $= M$
Velocity $= v$
Mechanical equivalent of heat $= J$
Since the gas flow is suddenly stopped, we will consider it to be an adiabatic process.
Change in energy $Δ$ K.E. $= \frac{1}{2} {mv}^{2} - \frac{1}{2} M .0$
$= \frac{1}{2} {mv}^{2}$
Work done for adiabatic process,
$W = \frac{nR Δ T}{γ - 1} W = \frac{P_{1} V_{1} - P_{2} V_{2}}{γ - 1}$
$Δ U = Δ W [for adiabatic process]$
$\frac{1}{2} {mv}^{2} = J \frac{nR Δ T}{γ - 1}$
$n = J \frac{m}{M} [$ no.of moles. $]$
$\frac{1}{2} {mv}^{2} = J \frac{mR Δ T}{M (γ - 1)}$
$\frac{v^{2}}{2} = \frac{JR Δ T}{M (γ - 1)}$
$Δ T = \frac{{Mv}^{2} (γ - 1)}{2 RJ}$

AP EAMCET(Medical)-2006

Thermodynamics

148508 During an experiment, an ideal gas is found to obey an additional law $V^{2} =$ constant. The gas is initially at temperature $T$ and volume. $V$ . The temperature of the gas will be following, when it expands to a volume $2 V$ ?

1

\sqrt{2} T

2

\sqrt{4} T

3

\sqrt{6} T

4

\sqrt{5} T

Explanation:

A Given,
$\begin{array}{ll} We know {VP}^{2} = constant \\ Again PV = nRT \\ P = \frac{nRT}{V} \end{array}$
Put in equation (i), We get,
$∴ \frac{(nRT)^{2}}{V} = constant$
$\frac{T^{2}}{V} = constant$
Thus volume $V$ when expanded to $2 V$ , temperature $T_{2}$
$\frac{T_{2}^{2}}{T_{1}^{2}} = \frac{V_{2}}{V_{1}} = \frac{2 V}{V}$
$T_{2} = \sqrt{2} T_{1}$

CG PET- 2014

Thermodynamics

148509 Two containers of equal volume contain the same gas at pressure $P_{1}$ and $P_{2}$ and absolute temperature $T_{1}$ and $T_{2}$ respectively. On joining the vessels the gas reaches a common pressure $P$ and common temperature $T$ . The ratio $P / T$ is equal to

1

\frac{P_{1}}{T_{1}} + \frac{P_{2}}{T_{2}}

2

\frac{P_{1} T_{1} + P_{2} T_{2}}{{(T_{1} + T_{2})}^{2}}

3

\frac{P_{1} T_{2} + P_{2} T_{1}}{{(T_{1} + T_{2})}^{2}}

4

\frac{P_{1}}{2 T_{1}} + \frac{P_{2}}{2 T_{2}}

Explanation:

D From ideal gas-
$PV = mRT$
$∴ m = \frac{PV}{RT}$
$∴$ Number of moles of gas in first container-
$m_{1} = \frac{P_{1} V}{{RT}_{1}}$
Number of moles of gas in second container-
$m_{2} = \frac{P_{2} V}{{RT}_{2}}$
Number of moles in containers when joined with each other.
$m = \frac{PV}{RT}$
But,
$m = m_{1} + m_{2}$
$\frac{P (2 V)}{RT} = \frac{P_{1} V}{{RT}_{1}} + \frac{P_{2} V}{{RT}_{2}}$
$\frac{2 P}{T} = \frac{P_{1}}{T_{1}} + \frac{P_{2}}{T_{2}}$
$\frac{P}{T} = \frac{P_{1}}{2 T_{1}} + \frac{P_{2}}{2 T_{2}}$

CG PET- 2006

Thermodynamics

148510 2 moles of an ideal mono-atomic gas is carried from a state $(P_{0}, V_{0})$ to state $(2 P_{0}, 2 V_{0})$ along a straight line path in a $P - V$ diagram. The amount of heat absorbed by the gas in the process is given by

1

3 P_{0} V_{0}

2

\frac{9}{2} P_{0} V_{0}

3

6 P_{0} V_{0}

4

\frac{3}{2} P_{0} V_{0}

Explanation:

C
Area under the curve given work done in $P - v$ graph
$W = V_{0} P_{0} + \frac{V_{0} P_{0}}{2}$
$W = \frac{3}{2} P_{0} V_{0}$
$Δ u = {nC}_{v} (T_{B} - T_{A})$
$= \frac{3 nR}{2} (\frac{2 P_{0} \times 2 V_{0}}{nR} - \frac{P_{0} V_{0}}{nR})$
$= \frac{9}{2} P_{0} V_{0}$
$Q = U + W$
$[according to first law of thermodynamic ]$
$Q = \frac{9}{2} P_{0} V_{0} + \frac{3}{2} P_{0} V_{0}$
$Q = 6 P_{0} V_{0}$

WB JEE 2017

Thermodynamics

148512 If the coefficient of superficial expansion is $x$ times the coefficient of cubical expansion, then the value of $x$ is

1 2

2

\frac{3}{2}

3

\frac{2}{3}

4

\frac{1}{2}

Explanation:

B Given that,
Coefficient of superficial expansion $= γ$
As we know that coefficient of cubical expansion is related to coefficient of superficial expansion.
The relation between cubical expansion and superficial expansion $β$ is given as
$\frac{β}{2} = \frac{γ}{3}$
$or γ = \frac{3 β}{2}$
It is given that in question coefficient of the cubical expansion is increased by the $x$ times of the coefficient of superficial expansion i.e.
$γ = β x$
From equation (i) and (ii),
$x = \frac{3}{2}$
$x = 1.5$

SCRA-2015

Thermodynamics

148513 ' $m$ ' grams of a gas of molecular weight $M$ is flowing in an isolated tube velocity $v$ . If the gas flow is suddenly stopped the rise in the temperature is : $(γ =$ ratio of specific heats, $R =$ universal gas constant, $J =$ mechanical equivalent of heat).

1

\frac{{Mv}^{2} (γ - 1)}{2 RJ}

2

\frac{m}{M} \frac{v^{2} (γ - 1)}{2 RJ}

3

\frac{m v^{2} γ}{2 RJ}

4

\frac{{Mv}^{2} γ}{2 RJ}

Explanation:

A Given that,
Mass of gas $= m$
Molecular weight $= M$
Velocity $= v$
Mechanical equivalent of heat $= J$
Since the gas flow is suddenly stopped, we will consider it to be an adiabatic process.
Change in energy $Δ$ K.E. $= \frac{1}{2} {mv}^{2} - \frac{1}{2} M .0$
$= \frac{1}{2} {mv}^{2}$
Work done for adiabatic process,
$W = \frac{nR Δ T}{γ - 1} W = \frac{P_{1} V_{1} - P_{2} V_{2}}{γ - 1}$
$Δ U = Δ W [for adiabatic process]$
$\frac{1}{2} {mv}^{2} = J \frac{nR Δ T}{γ - 1}$
$n = J \frac{m}{M} [$ no.of moles. $]$
$\frac{1}{2} {mv}^{2} = J \frac{mR Δ T}{M (γ - 1)}$
$\frac{v^{2}}{2} = \frac{JR Δ T}{M (γ - 1)}$
$Δ T = \frac{{Mv}^{2} (γ - 1)}{2 RJ}$

AP EAMCET(Medical)-2006

Thermodynamics

148508 During an experiment, an ideal gas is found to obey an additional law $V^{2} =$ constant. The gas is initially at temperature $T$ and volume. $V$ . The temperature of the gas will be following, when it expands to a volume $2 V$ ?

1

\sqrt{2} T

2

\sqrt{4} T

3

\sqrt{6} T

4

\sqrt{5} T

Explanation:

A Given,
$\begin{array}{ll} We know {VP}^{2} = constant \\ Again PV = nRT \\ P = \frac{nRT}{V} \end{array}$
Put in equation (i), We get,
$∴ \frac{(nRT)^{2}}{V} = constant$
$\frac{T^{2}}{V} = constant$
Thus volume $V$ when expanded to $2 V$ , temperature $T_{2}$
$\frac{T_{2}^{2}}{T_{1}^{2}} = \frac{V_{2}}{V_{1}} = \frac{2 V}{V}$
$T_{2} = \sqrt{2} T_{1}$

CG PET- 2014

Thermodynamics

148509 Two containers of equal volume contain the same gas at pressure $P_{1}$ and $P_{2}$ and absolute temperature $T_{1}$ and $T_{2}$ respectively. On joining the vessels the gas reaches a common pressure $P$ and common temperature $T$ . The ratio $P / T$ is equal to

1

\frac{P_{1}}{T_{1}} + \frac{P_{2}}{T_{2}}

2

\frac{P_{1} T_{1} + P_{2} T_{2}}{{(T_{1} + T_{2})}^{2}}

3

\frac{P_{1} T_{2} + P_{2} T_{1}}{{(T_{1} + T_{2})}^{2}}

4

\frac{P_{1}}{2 T_{1}} + \frac{P_{2}}{2 T_{2}}

Explanation:

D From ideal gas-
$PV = mRT$
$∴ m = \frac{PV}{RT}$
$∴$ Number of moles of gas in first container-
$m_{1} = \frac{P_{1} V}{{RT}_{1}}$
Number of moles of gas in second container-
$m_{2} = \frac{P_{2} V}{{RT}_{2}}$
Number of moles in containers when joined with each other.
$m = \frac{PV}{RT}$
But,
$m = m_{1} + m_{2}$
$\frac{P (2 V)}{RT} = \frac{P_{1} V}{{RT}_{1}} + \frac{P_{2} V}{{RT}_{2}}$
$\frac{2 P}{T} = \frac{P_{1}}{T_{1}} + \frac{P_{2}}{T_{2}}$
$\frac{P}{T} = \frac{P_{1}}{2 T_{1}} + \frac{P_{2}}{2 T_{2}}$

CG PET- 2006

Thermodynamics

148510 2 moles of an ideal mono-atomic gas is carried from a state $(P_{0}, V_{0})$ to state $(2 P_{0}, 2 V_{0})$ along a straight line path in a $P - V$ diagram. The amount of heat absorbed by the gas in the process is given by

1

3 P_{0} V_{0}

2

\frac{9}{2} P_{0} V_{0}

3

6 P_{0} V_{0}

4

\frac{3}{2} P_{0} V_{0}

Explanation:

C
Area under the curve given work done in $P - v$ graph
$W = V_{0} P_{0} + \frac{V_{0} P_{0}}{2}$
$W = \frac{3}{2} P_{0} V_{0}$
$Δ u = {nC}_{v} (T_{B} - T_{A})$
$= \frac{3 nR}{2} (\frac{2 P_{0} \times 2 V_{0}}{nR} - \frac{P_{0} V_{0}}{nR})$
$= \frac{9}{2} P_{0} V_{0}$
$Q = U + W$
$[according to first law of thermodynamic ]$
$Q = \frac{9}{2} P_{0} V_{0} + \frac{3}{2} P_{0} V_{0}$
$Q = 6 P_{0} V_{0}$

WB JEE 2017

Thermodynamics

148512 If the coefficient of superficial expansion is $x$ times the coefficient of cubical expansion, then the value of $x$ is

1 2

2

\frac{3}{2}

3

\frac{2}{3}

4

\frac{1}{2}

Explanation:

B Given that,
Coefficient of superficial expansion $= γ$
As we know that coefficient of cubical expansion is related to coefficient of superficial expansion.
The relation between cubical expansion and superficial expansion $β$ is given as
$\frac{β}{2} = \frac{γ}{3}$
$or γ = \frac{3 β}{2}$
It is given that in question coefficient of the cubical expansion is increased by the $x$ times of the coefficient of superficial expansion i.e.
$γ = β x$
From equation (i) and (ii),
$x = \frac{3}{2}$
$x = 1.5$

SCRA-2015

Thermodynamics

148513 ' $m$ ' grams of a gas of molecular weight $M$ is flowing in an isolated tube velocity $v$ . If the gas flow is suddenly stopped the rise in the temperature is : $(γ =$ ratio of specific heats, $R =$ universal gas constant, $J =$ mechanical equivalent of heat).

1

\frac{{Mv}^{2} (γ - 1)}{2 RJ}

2

\frac{m}{M} \frac{v^{2} (γ - 1)}{2 RJ}

3

\frac{m v^{2} γ}{2 RJ}

4

\frac{{Mv}^{2} γ}{2 RJ}

Explanation:

A Given that,
Mass of gas $= m$
Molecular weight $= M$
Velocity $= v$
Mechanical equivalent of heat $= J$
Since the gas flow is suddenly stopped, we will consider it to be an adiabatic process.
Change in energy $Δ$ K.E. $= \frac{1}{2} {mv}^{2} - \frac{1}{2} M .0$
$= \frac{1}{2} {mv}^{2}$
Work done for adiabatic process,
$W = \frac{nR Δ T}{γ - 1} W = \frac{P_{1} V_{1} - P_{2} V_{2}}{γ - 1}$
$Δ U = Δ W [for adiabatic process]$
$\frac{1}{2} {mv}^{2} = J \frac{nR Δ T}{γ - 1}$
$n = J \frac{m}{M} [$ no.of moles. $]$
$\frac{1}{2} {mv}^{2} = J \frac{mR Δ T}{M (γ - 1)}$
$\frac{v^{2}}{2} = \frac{JR Δ T}{M (γ - 1)}$
$Δ T = \frac{{Mv}^{2} (γ - 1)}{2 RJ}$

AP EAMCET(Medical)-2006