QBManager

Kinetic Theory of Gases

139354 The specific heat of helium at constant volume is $12.6 J {mol}^{- 1} K^{- 1}$ . The specific heat of helium at constant pressure in $J {mol}^{- 1} K^{- 1}$ is about ( Assume the temperature of the gas is moderate, universal gas constant, $R = 8.314 J {mol}^{- 1} K^{- 1}$ )

1 12.6

2 16.8

3 18.9

4 21

Explanation:

D Given that,
$C_{v} = 12.6 J {mol}^{- 1} k^{- 1}$
$C_{p} =$ ?
$R = 8.314 J {mol}^{- 1} k^{- 1}$
We know that,
$C_{p} - C_{v} = R$
$C_{p} = C_{v} + R$
$= 12.6 + 8.314$
$C_{p} = 20.9 \approx 21$

TS- EAMCET-10.09.2020

Kinetic Theory of Gases

139355 When two moles of oxygen is heated from $0^{\circ} C$ to $10^{\circ} C$ at constant volume, its internal energy changes by $420 J$ . What is the molar specific heat of oxygen at constant volume?

1

5.75 {JK}^{- 1} {mol}^{- 1}

2

10.5 {JK}^{- 1} {mol}^{- 1}

3

21 {JK}^{- 1} {mol}^{- 1}

4

42 {JK}^{- 1} {mol}^{- 1}

Explanation:

C Given that,
No of mole $(n) = 2$
Internal energy $(Δ U) = 420 J$
$Δ T = 10^{\circ} C - 0^{\circ} C = 10^{\circ} C$
Molar specific heat at constant volume $(C_{V}) =$ ?
We know that,
$Δ U = nCv Δ T$
$C_{V} = \frac{Δ U}{n Δ T}$
$= \frac{420}{2 \times 10}$
$C_{V} = 21 J K^{- 1} {mol}^{- 1}$

AMU-2009

Kinetic Theory of Gases

139356 If for a gas, $\frac{R}{C_{v}} = 0.67$ , this gas is made up of molecules which are,

1 diatomic

2 mixture of diatomic and polyatomic molecules

3 Monoatomic

4 Polyatomic

Explanation:

C Given that
$\frac{R}{C_{v}} = 0.67$
We know that,
$C_{v} = \frac{R}{γ - 1}$
$γ - 1 = \frac{R}{C_{v}}$
$γ - 1 = 0.67$
$γ = 1.67$
Hence, the gas is monoatomic.

MHT-CET 2020

Kinetic Theory of Gases

139357 One mole of an ideal monatomic gas undergoes a process described by the equation ${PV}^{3} =$ constant. The heat capacity of the gas during this process is

1

\frac{3}{2} R

2

\frac{5}{2} R

3

2 R

4

R

Explanation:

D Process describe by the equation ${PV}^{3} =$ Constant
For a polytropic process-
${PV}^{x} =$ Constant
$C = C_{v} + \frac{R}{1 - x}$
$C = \frac{f}{2} R + \frac{R}{1 - x}$
Here, $x = 3$
$f = degree of freedom = 3$
Hence, $C = \frac{3}{2} R + \frac{R}{1 - 3}$
$C = \frac{3 R}{2} - \frac{R}{2}$
$C = \frac{2 R}{2}$
$C = R$

NEET-2016

Kinetic Theory of Gases

139358 The amount of heat energy required to raise the temperature of $1 g$ of helium at NTP, from $T_{1} K$ to $T_{2} K$ is

1

\frac{3}{8} N_{a} K_{B} (T_{2} - T_{1})

2

\frac{3}{2} N_{a} K_{B} (T_{2} - T_{1})

3

\frac{3}{4} N_{a} K_{B} (T_{2} - T_{1})

4

\frac{3}{4} N_{a} K_{B} (\frac{T_{2}}{T_{1}})

Explanation:

A Given that,
Mass of helium $(m) = 1 g$
Molecular mass of helium $(M) = 4$
$Δ T = T_{2} - T_{1}$
We know that,
$Heat energy (Q) = \frac{3}{2} {nN}_{a} K_{B} Δ T$
$= \frac{3}{2} \frac{m}{M} N_{a} K_{B} (T_{2} - T_{1})$
$= \frac{3}{2} (\frac{1}{4}) N_{a} K_{B} (T_{2} - T_{1})$
$= \frac{3}{8} N_{a} K_{B} (T_{2} - T_{1})$

NEET-2013

Kinetic Theory of Gases

139354 The specific heat of helium at constant volume is $12.6 J {mol}^{- 1} K^{- 1}$ . The specific heat of helium at constant pressure in $J {mol}^{- 1} K^{- 1}$ is about ( Assume the temperature of the gas is moderate, universal gas constant, $R = 8.314 J {mol}^{- 1} K^{- 1}$ )

1 12.6

2 16.8

3 18.9

4 21

Explanation:

D Given that,
$C_{v} = 12.6 J {mol}^{- 1} k^{- 1}$
$C_{p} =$ ?
$R = 8.314 J {mol}^{- 1} k^{- 1}$
We know that,
$C_{p} - C_{v} = R$
$C_{p} = C_{v} + R$
$= 12.6 + 8.314$
$C_{p} = 20.9 \approx 21$

TS- EAMCET-10.09.2020

Kinetic Theory of Gases

139355 When two moles of oxygen is heated from $0^{\circ} C$ to $10^{\circ} C$ at constant volume, its internal energy changes by $420 J$ . What is the molar specific heat of oxygen at constant volume?

1

5.75 {JK}^{- 1} {mol}^{- 1}

2

10.5 {JK}^{- 1} {mol}^{- 1}

3

21 {JK}^{- 1} {mol}^{- 1}

4

42 {JK}^{- 1} {mol}^{- 1}

Explanation:

C Given that,
No of mole $(n) = 2$
Internal energy $(Δ U) = 420 J$
$Δ T = 10^{\circ} C - 0^{\circ} C = 10^{\circ} C$
Molar specific heat at constant volume $(C_{V}) =$ ?
We know that,
$Δ U = nCv Δ T$
$C_{V} = \frac{Δ U}{n Δ T}$
$= \frac{420}{2 \times 10}$
$C_{V} = 21 J K^{- 1} {mol}^{- 1}$

AMU-2009

Kinetic Theory of Gases

139356 If for a gas, $\frac{R}{C_{v}} = 0.67$ , this gas is made up of molecules which are,

1 diatomic

2 mixture of diatomic and polyatomic molecules

3 Monoatomic

4 Polyatomic

Explanation:

C Given that
$\frac{R}{C_{v}} = 0.67$
We know that,
$C_{v} = \frac{R}{γ - 1}$
$γ - 1 = \frac{R}{C_{v}}$
$γ - 1 = 0.67$
$γ = 1.67$
Hence, the gas is monoatomic.

MHT-CET 2020

Kinetic Theory of Gases

139357 One mole of an ideal monatomic gas undergoes a process described by the equation ${PV}^{3} =$ constant. The heat capacity of the gas during this process is

1

\frac{3}{2} R

2

\frac{5}{2} R

3

2 R

4

R

Explanation:

D Process describe by the equation ${PV}^{3} =$ Constant
For a polytropic process-
${PV}^{x} =$ Constant
$C = C_{v} + \frac{R}{1 - x}$
$C = \frac{f}{2} R + \frac{R}{1 - x}$
Here, $x = 3$
$f = degree of freedom = 3$
Hence, $C = \frac{3}{2} R + \frac{R}{1 - 3}$
$C = \frac{3 R}{2} - \frac{R}{2}$
$C = \frac{2 R}{2}$
$C = R$

NEET-2016

Kinetic Theory of Gases

139358 The amount of heat energy required to raise the temperature of $1 g$ of helium at NTP, from $T_{1} K$ to $T_{2} K$ is

1

\frac{3}{8} N_{a} K_{B} (T_{2} - T_{1})

2

\frac{3}{2} N_{a} K_{B} (T_{2} - T_{1})

3

\frac{3}{4} N_{a} K_{B} (T_{2} - T_{1})

4

\frac{3}{4} N_{a} K_{B} (\frac{T_{2}}{T_{1}})

Explanation:

A Given that,
Mass of helium $(m) = 1 g$
Molecular mass of helium $(M) = 4$
$Δ T = T_{2} - T_{1}$
We know that,
$Heat energy (Q) = \frac{3}{2} {nN}_{a} K_{B} Δ T$
$= \frac{3}{2} \frac{m}{M} N_{a} K_{B} (T_{2} - T_{1})$
$= \frac{3}{2} (\frac{1}{4}) N_{a} K_{B} (T_{2} - T_{1})$
$= \frac{3}{8} N_{a} K_{B} (T_{2} - T_{1})$

NEET-2013

Kinetic Theory of Gases

139354 The specific heat of helium at constant volume is $12.6 J {mol}^{- 1} K^{- 1}$ . The specific heat of helium at constant pressure in $J {mol}^{- 1} K^{- 1}$ is about ( Assume the temperature of the gas is moderate, universal gas constant, $R = 8.314 J {mol}^{- 1} K^{- 1}$ )

1 12.6

2 16.8

3 18.9

4 21

Explanation:

D Given that,
$C_{v} = 12.6 J {mol}^{- 1} k^{- 1}$
$C_{p} =$ ?
$R = 8.314 J {mol}^{- 1} k^{- 1}$
We know that,
$C_{p} - C_{v} = R$
$C_{p} = C_{v} + R$
$= 12.6 + 8.314$
$C_{p} = 20.9 \approx 21$

TS- EAMCET-10.09.2020

Kinetic Theory of Gases

139355 When two moles of oxygen is heated from $0^{\circ} C$ to $10^{\circ} C$ at constant volume, its internal energy changes by $420 J$ . What is the molar specific heat of oxygen at constant volume?

1

5.75 {JK}^{- 1} {mol}^{- 1}

2

10.5 {JK}^{- 1} {mol}^{- 1}

3

21 {JK}^{- 1} {mol}^{- 1}

4

42 {JK}^{- 1} {mol}^{- 1}

Explanation:

C Given that,
No of mole $(n) = 2$
Internal energy $(Δ U) = 420 J$
$Δ T = 10^{\circ} C - 0^{\circ} C = 10^{\circ} C$
Molar specific heat at constant volume $(C_{V}) =$ ?
We know that,
$Δ U = nCv Δ T$
$C_{V} = \frac{Δ U}{n Δ T}$
$= \frac{420}{2 \times 10}$
$C_{V} = 21 J K^{- 1} {mol}^{- 1}$

AMU-2009

Kinetic Theory of Gases

139356 If for a gas, $\frac{R}{C_{v}} = 0.67$ , this gas is made up of molecules which are,

1 diatomic

2 mixture of diatomic and polyatomic molecules

3 Monoatomic

4 Polyatomic

Explanation:

C Given that
$\frac{R}{C_{v}} = 0.67$
We know that,
$C_{v} = \frac{R}{γ - 1}$
$γ - 1 = \frac{R}{C_{v}}$
$γ - 1 = 0.67$
$γ = 1.67$
Hence, the gas is monoatomic.

MHT-CET 2020

Kinetic Theory of Gases

139357 One mole of an ideal monatomic gas undergoes a process described by the equation ${PV}^{3} =$ constant. The heat capacity of the gas during this process is

1

\frac{3}{2} R

2

\frac{5}{2} R

3

2 R

4

R

Explanation:

D Process describe by the equation ${PV}^{3} =$ Constant
For a polytropic process-
${PV}^{x} =$ Constant
$C = C_{v} + \frac{R}{1 - x}$
$C = \frac{f}{2} R + \frac{R}{1 - x}$
Here, $x = 3$
$f = degree of freedom = 3$
Hence, $C = \frac{3}{2} R + \frac{R}{1 - 3}$
$C = \frac{3 R}{2} - \frac{R}{2}$
$C = \frac{2 R}{2}$
$C = R$

NEET-2016

Kinetic Theory of Gases

139358 The amount of heat energy required to raise the temperature of $1 g$ of helium at NTP, from $T_{1} K$ to $T_{2} K$ is

1

\frac{3}{8} N_{a} K_{B} (T_{2} - T_{1})

2

\frac{3}{2} N_{a} K_{B} (T_{2} - T_{1})

3

\frac{3}{4} N_{a} K_{B} (T_{2} - T_{1})

4

\frac{3}{4} N_{a} K_{B} (\frac{T_{2}}{T_{1}})

Explanation:

A Given that,
Mass of helium $(m) = 1 g$
Molecular mass of helium $(M) = 4$
$Δ T = T_{2} - T_{1}$
We know that,
$Heat energy (Q) = \frac{3}{2} {nN}_{a} K_{B} Δ T$
$= \frac{3}{2} \frac{m}{M} N_{a} K_{B} (T_{2} - T_{1})$
$= \frac{3}{2} (\frac{1}{4}) N_{a} K_{B} (T_{2} - T_{1})$
$= \frac{3}{8} N_{a} K_{B} (T_{2} - T_{1})$

NEET-2013

Kinetic Theory of Gases

139354 The specific heat of helium at constant volume is $12.6 J {mol}^{- 1} K^{- 1}$ . The specific heat of helium at constant pressure in $J {mol}^{- 1} K^{- 1}$ is about ( Assume the temperature of the gas is moderate, universal gas constant, $R = 8.314 J {mol}^{- 1} K^{- 1}$ )

1 12.6

2 16.8

3 18.9

4 21

Explanation:

D Given that,
$C_{v} = 12.6 J {mol}^{- 1} k^{- 1}$
$C_{p} =$ ?
$R = 8.314 J {mol}^{- 1} k^{- 1}$
We know that,
$C_{p} - C_{v} = R$
$C_{p} = C_{v} + R$
$= 12.6 + 8.314$
$C_{p} = 20.9 \approx 21$

TS- EAMCET-10.09.2020

Kinetic Theory of Gases

139355 When two moles of oxygen is heated from $0^{\circ} C$ to $10^{\circ} C$ at constant volume, its internal energy changes by $420 J$ . What is the molar specific heat of oxygen at constant volume?

1

5.75 {JK}^{- 1} {mol}^{- 1}

2

10.5 {JK}^{- 1} {mol}^{- 1}

3

21 {JK}^{- 1} {mol}^{- 1}

4

42 {JK}^{- 1} {mol}^{- 1}

Explanation:

C Given that,
No of mole $(n) = 2$
Internal energy $(Δ U) = 420 J$
$Δ T = 10^{\circ} C - 0^{\circ} C = 10^{\circ} C$
Molar specific heat at constant volume $(C_{V}) =$ ?
We know that,
$Δ U = nCv Δ T$
$C_{V} = \frac{Δ U}{n Δ T}$
$= \frac{420}{2 \times 10}$
$C_{V} = 21 J K^{- 1} {mol}^{- 1}$

AMU-2009

Kinetic Theory of Gases

139356 If for a gas, $\frac{R}{C_{v}} = 0.67$ , this gas is made up of molecules which are,

1 diatomic

2 mixture of diatomic and polyatomic molecules

3 Monoatomic

4 Polyatomic

Explanation:

C Given that
$\frac{R}{C_{v}} = 0.67$
We know that,
$C_{v} = \frac{R}{γ - 1}$
$γ - 1 = \frac{R}{C_{v}}$
$γ - 1 = 0.67$
$γ = 1.67$
Hence, the gas is monoatomic.

MHT-CET 2020

Kinetic Theory of Gases

139357 One mole of an ideal monatomic gas undergoes a process described by the equation ${PV}^{3} =$ constant. The heat capacity of the gas during this process is

1

\frac{3}{2} R

2

\frac{5}{2} R

3

2 R

4

R

Explanation:

D Process describe by the equation ${PV}^{3} =$ Constant
For a polytropic process-
${PV}^{x} =$ Constant
$C = C_{v} + \frac{R}{1 - x}$
$C = \frac{f}{2} R + \frac{R}{1 - x}$
Here, $x = 3$
$f = degree of freedom = 3$
Hence, $C = \frac{3}{2} R + \frac{R}{1 - 3}$
$C = \frac{3 R}{2} - \frac{R}{2}$
$C = \frac{2 R}{2}$
$C = R$

NEET-2016

Kinetic Theory of Gases

139358 The amount of heat energy required to raise the temperature of $1 g$ of helium at NTP, from $T_{1} K$ to $T_{2} K$ is

1

\frac{3}{8} N_{a} K_{B} (T_{2} - T_{1})

2

\frac{3}{2} N_{a} K_{B} (T_{2} - T_{1})

3

\frac{3}{4} N_{a} K_{B} (T_{2} - T_{1})

4

\frac{3}{4} N_{a} K_{B} (\frac{T_{2}}{T_{1}})

Explanation:

A Given that,
Mass of helium $(m) = 1 g$
Molecular mass of helium $(M) = 4$
$Δ T = T_{2} - T_{1}$
We know that,
$Heat energy (Q) = \frac{3}{2} {nN}_{a} K_{B} Δ T$
$= \frac{3}{2} \frac{m}{M} N_{a} K_{B} (T_{2} - T_{1})$
$= \frac{3}{2} (\frac{1}{4}) N_{a} K_{B} (T_{2} - T_{1})$
$= \frac{3}{8} N_{a} K_{B} (T_{2} - T_{1})$

NEET-2013

Kinetic Theory of Gases

139354 The specific heat of helium at constant volume is $12.6 J {mol}^{- 1} K^{- 1}$ . The specific heat of helium at constant pressure in $J {mol}^{- 1} K^{- 1}$ is about ( Assume the temperature of the gas is moderate, universal gas constant, $R = 8.314 J {mol}^{- 1} K^{- 1}$ )

1 12.6

2 16.8

3 18.9

4 21

Explanation:

D Given that,
$C_{v} = 12.6 J {mol}^{- 1} k^{- 1}$
$C_{p} =$ ?
$R = 8.314 J {mol}^{- 1} k^{- 1}$
We know that,
$C_{p} - C_{v} = R$
$C_{p} = C_{v} + R$
$= 12.6 + 8.314$
$C_{p} = 20.9 \approx 21$

TS- EAMCET-10.09.2020

Kinetic Theory of Gases

139355 When two moles of oxygen is heated from $0^{\circ} C$ to $10^{\circ} C$ at constant volume, its internal energy changes by $420 J$ . What is the molar specific heat of oxygen at constant volume?

1

5.75 {JK}^{- 1} {mol}^{- 1}

2

10.5 {JK}^{- 1} {mol}^{- 1}

3

21 {JK}^{- 1} {mol}^{- 1}

4

42 {JK}^{- 1} {mol}^{- 1}

Explanation:

C Given that,
No of mole $(n) = 2$
Internal energy $(Δ U) = 420 J$
$Δ T = 10^{\circ} C - 0^{\circ} C = 10^{\circ} C$
Molar specific heat at constant volume $(C_{V}) =$ ?
We know that,
$Δ U = nCv Δ T$
$C_{V} = \frac{Δ U}{n Δ T}$
$= \frac{420}{2 \times 10}$
$C_{V} = 21 J K^{- 1} {mol}^{- 1}$

AMU-2009

Kinetic Theory of Gases

139356 If for a gas, $\frac{R}{C_{v}} = 0.67$ , this gas is made up of molecules which are,

1 diatomic

2 mixture of diatomic and polyatomic molecules

3 Monoatomic

4 Polyatomic

Explanation:

C Given that
$\frac{R}{C_{v}} = 0.67$
We know that,
$C_{v} = \frac{R}{γ - 1}$
$γ - 1 = \frac{R}{C_{v}}$
$γ - 1 = 0.67$
$γ = 1.67$
Hence, the gas is monoatomic.

MHT-CET 2020

Kinetic Theory of Gases

139357 One mole of an ideal monatomic gas undergoes a process described by the equation ${PV}^{3} =$ constant. The heat capacity of the gas during this process is

1

\frac{3}{2} R

2

\frac{5}{2} R

3

2 R

4

R

Explanation:

D Process describe by the equation ${PV}^{3} =$ Constant
For a polytropic process-
${PV}^{x} =$ Constant
$C = C_{v} + \frac{R}{1 - x}$
$C = \frac{f}{2} R + \frac{R}{1 - x}$
Here, $x = 3$
$f = degree of freedom = 3$
Hence, $C = \frac{3}{2} R + \frac{R}{1 - 3}$
$C = \frac{3 R}{2} - \frac{R}{2}$
$C = \frac{2 R}{2}$
$C = R$

NEET-2016

Kinetic Theory of Gases

139358 The amount of heat energy required to raise the temperature of $1 g$ of helium at NTP, from $T_{1} K$ to $T_{2} K$ is

1

\frac{3}{8} N_{a} K_{B} (T_{2} - T_{1})

2

\frac{3}{2} N_{a} K_{B} (T_{2} - T_{1})

3

\frac{3}{4} N_{a} K_{B} (T_{2} - T_{1})

4

\frac{3}{4} N_{a} K_{B} (\frac{T_{2}}{T_{1}})

Explanation:

A Given that,
Mass of helium $(m) = 1 g$
Molecular mass of helium $(M) = 4$
$Δ T = T_{2} - T_{1}$
We know that,
$Heat energy (Q) = \frac{3}{2} {nN}_{a} K_{B} Δ T$
$= \frac{3}{2} \frac{m}{M} N_{a} K_{B} (T_{2} - T_{1})$
$= \frac{3}{2} (\frac{1}{4}) N_{a} K_{B} (T_{2} - T_{1})$
$= \frac{3}{8} N_{a} K_{B} (T_{2} - T_{1})$

NEET-2013

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