QBManager

Kinetic Theory of Gases

139049 The volume of an ideal gas with adiabatic exponent $γ$ varies according to law $V = \frac{α}{T}$ where $α$ is a constant. If the temperature of gas increases by $Δ T$ , then the amount of heat absorbed by the gas is

1

\frac{R Δ T}{γ - 1}

2

R Δ T (\frac{1 - γ}{2 - γ})

3

R Δ T (\frac{2 - γ}{γ - 1})

4

R Δ T

Explanation:

C Given, $V = \frac{α}{T}$ .
Where, $α =$ Constant
$VT = α$
$V . (PV) = α$
${PV}^{2} = α$
${PV}^{2} = Constant$
We know that, ${PV}^{x} =$ constant
Comparing equation (i) and (ii), we get-
$∴ x = 2$ [after comparing with equation (i)]
And molar heat capacity is given by-
$C = \frac{R}{γ - 1} + \frac{R}{1 - x}$
$C = \frac{R}{γ - 1} + \frac{R}{1 - 2} (∵ x = 2)$
$C = \frac{R}{γ - 1} - R = \frac{R - R (γ - 1)}{γ - 1}$
$C = \frac{R - R γ + R}{γ - 1} = \frac{2 R - R γ}{γ - 1} = \frac{R (2 - γ)}{γ - 1}$
$C = \frac{(2 - γ) R}{γ - 1}$
Since the temperature of gas increases by $Δ T$
Then, $Q = mC Δ T$
$= 1 \times R \frac{(2 - γ)}{γ - 1} \times Δ T = R Δ T (\frac{2 - γ}{γ - 1})$
Therefore, the amount of heat absorbed by the gas is $R Δ T (\frac{2 - γ}{γ - 1})$

SCRA-2015

Kinetic Theory of Gases

139051 What is the mass of 2 litres of nitrogen at 22.4 atmospheric pressure and $273 K$ ?

1

28 g

2

14 \times 22.4 g

3

56 g

4 None of these

Explanation:

C Given that, volume of $N_{2} = 2 L = 2 \times 10^{- 3} m^{3}$ Pressure $(P) = 22.4 atm = 22.4 \times 1.01 \times 10^{5} = 22.6 \times 10^{5}$ $N / m^{2}$ , temperature $(T) = 273 K$
According to an ideal gas equation-
$PV = nRT$
$n = \frac{PV}{RT} = \frac{22.6 \times 10^{5} \times 2 \times 10^{- 3}}{8.314 \times 273} = 2$
We know that,
$No. of mole (n) = \frac{Mass of gas}{Molar mass of gas}$
Or
$2 = \frac{m}{28}$
Hence, the mass of $N_{2}$ is $56 g$ .

J and K CET- 2005

Kinetic Theory of Gases

139052 70 cal of heat is required to raise the temperature of 2 moles of an ideal gas from $30^{\circ} C$ to $35^{\circ} C$ while the pressure of the gas is kept constant. The amount of the heat required to raise the temperature of the same gas through the same temperature range at constant volume is : (gas constant $R = 2$ $c a l / mol - K)$

1

70 cal

2

60 cal

3

50 cal

4

30 cal

Explanation:

C Given that, required heat $(Δ Q)_{P} = 70$ cal. $Δ T = 35^{\circ} C - 30^{\circ} C = 5^{\circ} C$
We know that, $C_{P} = \frac{(Δ Q)_{P}}{m Δ T} = \frac{70}{2 \times 5} = 7 cal K^{- 1} {mol}^{- 1}$
$∵ C_{P} - C_{V} = R$
Or $C_{V} = C_{P} - R$
$= 7 - 2 = 5 cal K^{- 1} {mol}^{- 1}$
At constant volume, $(Δ Q)_{V} = {mC}_{v} Δ T$
$= 2 \times 5 \times 5 = 50 cal .$

UPSEE - 2004

Kinetic Theory of Gases

139053 A sample of ideal monatomic gas is taken round the cycle $A B C A$ as shown in the figure. The work done during the cycle is

1 zero

2

3 PV

3 6PV

4

9 PV

Explanation:

B The work done during the cycle is given by area under $P - V$ diagram.
$∵$ Work done $=$ Area enclosed by the $△ ABC$
$= \frac{1}{2} \times AC \times BC$
$= \frac{1}{2} \times (3 V - V) (4 P - P)$
$= \frac{1}{2} \times 2 V \times 3 P$
$= 3 PV$

CG PET- 2010

Kinetic Theory of Gases

139049 The volume of an ideal gas with adiabatic exponent $γ$ varies according to law $V = \frac{α}{T}$ where $α$ is a constant. If the temperature of gas increases by $Δ T$ , then the amount of heat absorbed by the gas is

1

\frac{R Δ T}{γ - 1}

2

R Δ T (\frac{1 - γ}{2 - γ})

3

R Δ T (\frac{2 - γ}{γ - 1})

4

R Δ T

Explanation:

C Given, $V = \frac{α}{T}$ .
Where, $α =$ Constant
$VT = α$
$V . (PV) = α$
${PV}^{2} = α$
${PV}^{2} = Constant$
We know that, ${PV}^{x} =$ constant
Comparing equation (i) and (ii), we get-
$∴ x = 2$ [after comparing with equation (i)]
And molar heat capacity is given by-
$C = \frac{R}{γ - 1} + \frac{R}{1 - x}$
$C = \frac{R}{γ - 1} + \frac{R}{1 - 2} (∵ x = 2)$
$C = \frac{R}{γ - 1} - R = \frac{R - R (γ - 1)}{γ - 1}$
$C = \frac{R - R γ + R}{γ - 1} = \frac{2 R - R γ}{γ - 1} = \frac{R (2 - γ)}{γ - 1}$
$C = \frac{(2 - γ) R}{γ - 1}$
Since the temperature of gas increases by $Δ T$
Then, $Q = mC Δ T$
$= 1 \times R \frac{(2 - γ)}{γ - 1} \times Δ T = R Δ T (\frac{2 - γ}{γ - 1})$
Therefore, the amount of heat absorbed by the gas is $R Δ T (\frac{2 - γ}{γ - 1})$

SCRA-2015

Kinetic Theory of Gases

139051 What is the mass of 2 litres of nitrogen at 22.4 atmospheric pressure and $273 K$ ?

1

28 g

2

14 \times 22.4 g

3

56 g

4 None of these

Explanation:

C Given that, volume of $N_{2} = 2 L = 2 \times 10^{- 3} m^{3}$ Pressure $(P) = 22.4 atm = 22.4 \times 1.01 \times 10^{5} = 22.6 \times 10^{5}$ $N / m^{2}$ , temperature $(T) = 273 K$
According to an ideal gas equation-
$PV = nRT$
$n = \frac{PV}{RT} = \frac{22.6 \times 10^{5} \times 2 \times 10^{- 3}}{8.314 \times 273} = 2$
We know that,
$No. of mole (n) = \frac{Mass of gas}{Molar mass of gas}$
Or
$2 = \frac{m}{28}$
Hence, the mass of $N_{2}$ is $56 g$ .

J and K CET- 2005

Kinetic Theory of Gases

139052 70 cal of heat is required to raise the temperature of 2 moles of an ideal gas from $30^{\circ} C$ to $35^{\circ} C$ while the pressure of the gas is kept constant. The amount of the heat required to raise the temperature of the same gas through the same temperature range at constant volume is : (gas constant $R = 2$ $c a l / mol - K)$

1

70 cal

2

60 cal

3

50 cal

4

30 cal

Explanation:

C Given that, required heat $(Δ Q)_{P} = 70$ cal. $Δ T = 35^{\circ} C - 30^{\circ} C = 5^{\circ} C$
We know that, $C_{P} = \frac{(Δ Q)_{P}}{m Δ T} = \frac{70}{2 \times 5} = 7 cal K^{- 1} {mol}^{- 1}$
$∵ C_{P} - C_{V} = R$
Or $C_{V} = C_{P} - R$
$= 7 - 2 = 5 cal K^{- 1} {mol}^{- 1}$
At constant volume, $(Δ Q)_{V} = {mC}_{v} Δ T$
$= 2 \times 5 \times 5 = 50 cal .$

UPSEE - 2004

Kinetic Theory of Gases

139053 A sample of ideal monatomic gas is taken round the cycle $A B C A$ as shown in the figure. The work done during the cycle is

1 zero

2

3 PV

3 6PV

4

9 PV

Explanation:

B The work done during the cycle is given by area under $P - V$ diagram.
$∵$ Work done $=$ Area enclosed by the $△ ABC$
$= \frac{1}{2} \times AC \times BC$
$= \frac{1}{2} \times (3 V - V) (4 P - P)$
$= \frac{1}{2} \times 2 V \times 3 P$
$= 3 PV$

CG PET- 2010

Kinetic Theory of Gases

139049 The volume of an ideal gas with adiabatic exponent $γ$ varies according to law $V = \frac{α}{T}$ where $α$ is a constant. If the temperature of gas increases by $Δ T$ , then the amount of heat absorbed by the gas is

1

\frac{R Δ T}{γ - 1}

2

R Δ T (\frac{1 - γ}{2 - γ})

3

R Δ T (\frac{2 - γ}{γ - 1})

4

R Δ T

Explanation:

C Given, $V = \frac{α}{T}$ .
Where, $α =$ Constant
$VT = α$
$V . (PV) = α$
${PV}^{2} = α$
${PV}^{2} = Constant$
We know that, ${PV}^{x} =$ constant
Comparing equation (i) and (ii), we get-
$∴ x = 2$ [after comparing with equation (i)]
And molar heat capacity is given by-
$C = \frac{R}{γ - 1} + \frac{R}{1 - x}$
$C = \frac{R}{γ - 1} + \frac{R}{1 - 2} (∵ x = 2)$
$C = \frac{R}{γ - 1} - R = \frac{R - R (γ - 1)}{γ - 1}$
$C = \frac{R - R γ + R}{γ - 1} = \frac{2 R - R γ}{γ - 1} = \frac{R (2 - γ)}{γ - 1}$
$C = \frac{(2 - γ) R}{γ - 1}$
Since the temperature of gas increases by $Δ T$
Then, $Q = mC Δ T$
$= 1 \times R \frac{(2 - γ)}{γ - 1} \times Δ T = R Δ T (\frac{2 - γ}{γ - 1})$
Therefore, the amount of heat absorbed by the gas is $R Δ T (\frac{2 - γ}{γ - 1})$

SCRA-2015

Kinetic Theory of Gases

139051 What is the mass of 2 litres of nitrogen at 22.4 atmospheric pressure and $273 K$ ?

1

28 g

2

14 \times 22.4 g

3

56 g

4 None of these

Explanation:

C Given that, volume of $N_{2} = 2 L = 2 \times 10^{- 3} m^{3}$ Pressure $(P) = 22.4 atm = 22.4 \times 1.01 \times 10^{5} = 22.6 \times 10^{5}$ $N / m^{2}$ , temperature $(T) = 273 K$
According to an ideal gas equation-
$PV = nRT$
$n = \frac{PV}{RT} = \frac{22.6 \times 10^{5} \times 2 \times 10^{- 3}}{8.314 \times 273} = 2$
We know that,
$No. of mole (n) = \frac{Mass of gas}{Molar mass of gas}$
Or
$2 = \frac{m}{28}$
Hence, the mass of $N_{2}$ is $56 g$ .

J and K CET- 2005

Kinetic Theory of Gases

139052 70 cal of heat is required to raise the temperature of 2 moles of an ideal gas from $30^{\circ} C$ to $35^{\circ} C$ while the pressure of the gas is kept constant. The amount of the heat required to raise the temperature of the same gas through the same temperature range at constant volume is : (gas constant $R = 2$ $c a l / mol - K)$

1

70 cal

2

60 cal

3

50 cal

4

30 cal

Explanation:

C Given that, required heat $(Δ Q)_{P} = 70$ cal. $Δ T = 35^{\circ} C - 30^{\circ} C = 5^{\circ} C$
We know that, $C_{P} = \frac{(Δ Q)_{P}}{m Δ T} = \frac{70}{2 \times 5} = 7 cal K^{- 1} {mol}^{- 1}$
$∵ C_{P} - C_{V} = R$
Or $C_{V} = C_{P} - R$
$= 7 - 2 = 5 cal K^{- 1} {mol}^{- 1}$
At constant volume, $(Δ Q)_{V} = {mC}_{v} Δ T$
$= 2 \times 5 \times 5 = 50 cal .$

UPSEE - 2004

Kinetic Theory of Gases

139053 A sample of ideal monatomic gas is taken round the cycle $A B C A$ as shown in the figure. The work done during the cycle is

1 zero

2

3 PV

3 6PV

4

9 PV

Explanation:

B The work done during the cycle is given by area under $P - V$ diagram.
$∵$ Work done $=$ Area enclosed by the $△ ABC$
$= \frac{1}{2} \times AC \times BC$
$= \frac{1}{2} \times (3 V - V) (4 P - P)$
$= \frac{1}{2} \times 2 V \times 3 P$
$= 3 PV$

CG PET- 2010

Kinetic Theory of Gases

139049 The volume of an ideal gas with adiabatic exponent $γ$ varies according to law $V = \frac{α}{T}$ where $α$ is a constant. If the temperature of gas increases by $Δ T$ , then the amount of heat absorbed by the gas is

1

\frac{R Δ T}{γ - 1}

2

R Δ T (\frac{1 - γ}{2 - γ})

3

R Δ T (\frac{2 - γ}{γ - 1})

4

R Δ T

Explanation:

C Given, $V = \frac{α}{T}$ .
Where, $α =$ Constant
$VT = α$
$V . (PV) = α$
${PV}^{2} = α$
${PV}^{2} = Constant$
We know that, ${PV}^{x} =$ constant
Comparing equation (i) and (ii), we get-
$∴ x = 2$ [after comparing with equation (i)]
And molar heat capacity is given by-
$C = \frac{R}{γ - 1} + \frac{R}{1 - x}$
$C = \frac{R}{γ - 1} + \frac{R}{1 - 2} (∵ x = 2)$
$C = \frac{R}{γ - 1} - R = \frac{R - R (γ - 1)}{γ - 1}$
$C = \frac{R - R γ + R}{γ - 1} = \frac{2 R - R γ}{γ - 1} = \frac{R (2 - γ)}{γ - 1}$
$C = \frac{(2 - γ) R}{γ - 1}$
Since the temperature of gas increases by $Δ T$
Then, $Q = mC Δ T$
$= 1 \times R \frac{(2 - γ)}{γ - 1} \times Δ T = R Δ T (\frac{2 - γ}{γ - 1})$
Therefore, the amount of heat absorbed by the gas is $R Δ T (\frac{2 - γ}{γ - 1})$

SCRA-2015

Kinetic Theory of Gases

139051 What is the mass of 2 litres of nitrogen at 22.4 atmospheric pressure and $273 K$ ?

1

28 g

2

14 \times 22.4 g

3

56 g

4 None of these

Explanation:

C Given that, volume of $N_{2} = 2 L = 2 \times 10^{- 3} m^{3}$ Pressure $(P) = 22.4 atm = 22.4 \times 1.01 \times 10^{5} = 22.6 \times 10^{5}$ $N / m^{2}$ , temperature $(T) = 273 K$
According to an ideal gas equation-
$PV = nRT$
$n = \frac{PV}{RT} = \frac{22.6 \times 10^{5} \times 2 \times 10^{- 3}}{8.314 \times 273} = 2$
We know that,
$No. of mole (n) = \frac{Mass of gas}{Molar mass of gas}$
Or
$2 = \frac{m}{28}$
Hence, the mass of $N_{2}$ is $56 g$ .

J and K CET- 2005

Kinetic Theory of Gases

139052 70 cal of heat is required to raise the temperature of 2 moles of an ideal gas from $30^{\circ} C$ to $35^{\circ} C$ while the pressure of the gas is kept constant. The amount of the heat required to raise the temperature of the same gas through the same temperature range at constant volume is : (gas constant $R = 2$ $c a l / mol - K)$

1

70 cal

2

60 cal

3

50 cal

4

30 cal

Explanation:

C Given that, required heat $(Δ Q)_{P} = 70$ cal. $Δ T = 35^{\circ} C - 30^{\circ} C = 5^{\circ} C$
We know that, $C_{P} = \frac{(Δ Q)_{P}}{m Δ T} = \frac{70}{2 \times 5} = 7 cal K^{- 1} {mol}^{- 1}$
$∵ C_{P} - C_{V} = R$
Or $C_{V} = C_{P} - R$
$= 7 - 2 = 5 cal K^{- 1} {mol}^{- 1}$
At constant volume, $(Δ Q)_{V} = {mC}_{v} Δ T$
$= 2 \times 5 \times 5 = 50 cal .$

UPSEE - 2004

Kinetic Theory of Gases

139053 A sample of ideal monatomic gas is taken round the cycle $A B C A$ as shown in the figure. The work done during the cycle is

1 zero

2

3 PV

3 6PV

4

9 PV

Explanation:

B The work done during the cycle is given by area under $P - V$ diagram.
$∵$ Work done $=$ Area enclosed by the $△ ABC$
$= \frac{1}{2} \times AC \times BC$
$= \frac{1}{2} \times (3 V - V) (4 P - P)$
$= \frac{1}{2} \times 2 V \times 3 P$
$= 3 PV$

CG PET- 2010