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Kinetic Theory of Gases

138996 An ideal gas at temperature $T$ , pressure $p$ occupies a volume $V$ . If its temperature is halved and pressure doubled, what is its new volume?

1

\frac{V}{4}

2

\frac{V}{2}

3

V

4

2 V

Explanation:

A According to combined gas law equation.
$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$
Where $P_{1}, V_{1}, T_{1}$ are the pressure, volume and temperature of the gas in initial state And
$P_{2}, V_{2}, T_{2}$ are the pressure volume and temperature of gas in final state.
$V_{2} = \frac{P_{1} V_{1} T_{2}}{T_{1} P_{2}}$
Given, initial temperature $(T_{1}) = T$ , Pressure $(P_{1}) = P$ and volume $(V_{1}) = V$
Final temperature $(T_{2}) = \frac{T}{2}$
Final pressure $(P_{2}) = 2 P$
Then, $V_{2} =$ ?
$V_{2} = \frac{P \times V \times T}{T \times 2 P \times 2}$
$V_{2} = \frac{V}{4}$

Shift-1]

Kinetic Theory of Gases

138998 A cyclic process on 1 mole of an ideal gas is shown in the figure. The temperatures of the gas at points $A$ and $B$ are respectively $200 K$ and $400 K$ . A total of $200 J$ of heat is taken out from the system in the process. What is the work done by the gas in the process $B \to C$ ? (Let $R = 8.3 J K^{- 1} {mol}^{- 1}$ )

1

- 1660 J

2

1660 J

3

- 1860 J

4

+ 1860 J

Explanation:

C Given, $n = 1$ mole, $Δ Q = - 200 J, R = 8.3 J K^{- 1} {mol}^{- 1}$

In processes $C \to A$ volume is constant therefore work done zero.
In the cyclic processes change in internal energy will be zero then $dU = 0$
According to first law of thermodynamic -
$dQ = dU + dW$
$dQ = 0 + dW$
$dQ = W_{AB} + W_{BC} + W_{CA}$
$- 200 J = W_{AB} + W_{BC} + 0$
Then, $- 200 J = P (V_{B} - V_{A}) + W_{BC}$
$- 200 J = nR (T_{B} - T_{A}) + W_{BC}$
$- 200 J = 1 \times 8.3 (400 - 200) + W_{BC}$
$- 200 J = 8.3 \times 200 + W_{BC}$
$- 200 J = 1660.0 + W_{BC}$
Then $W_{BC} = - 200 J - 1660 J = - 1860 J$

Shift-I]

Kinetic Theory of Gases

138999 Molecular weight of oxygen is 32 amu. At STP, volume of $1 gm$ of oxygen is $700 {cm}^{3}$ . What is the value of gas constant $R$ ?

1

18.23 J {Mol}^{- 1} K^{- 1}

2

27.73 J {mol}^{- 1} K^{- 1}

3

37.97 J {mol}^{- 1} K^{- 1}

4

8.31 J {mol}^{- 1} K^{- 1}

Explanation:

D Given, molecular weight of $O_{2} = 32 amu$ At STP, $P = 1 atm = 1.013 \times 10^{5}, T = 273 K$
$∴ 1 gm of O_{2} - 700 {cm}^{3} (given)$
$32 gm of O_{2} - 700 \times 32 {cm}^{3}$
$700 \times 32 \times 10^{- 6} m^{3}$
For 1 mole of an ideal gas equation-
$PV = RT$
$R = \frac{PV}{T} = \frac{1.013 \times 10^{5} \times 700 \times 32 \times 10^{- 6}}{273}$
$R = 8.31 {mol}^{- 1} K^{- 1}$

J and K-CET-2019

Kinetic Theory of Gases

139000 A thermally insulated piston divides a container into a two compartments. The volume, temperature and pressure in the right compartment are $2 V, T$ and $2 P$ , while in the left compartment the corresponding parameters are $V, T$ and $P$ . Then the ratio of number of molecules in the right compartment to that in the left compartment is

1

1 : 1

2

2 : 1

3

4 : 1

4

1 : 4

Explanation:

C According to Ideal gas equation $PV = nRT$
For left compartment,
$PV = \frac{n_{1}}{N_{A}} RT$
(Where, $N_{A} =$ Avogadro No.)
For right compartment
$(2 P) (2 V) = \frac{n_{2}}{N_{A}} RT$
Dividing equation (i) by (ii), we get
$\frac{PV}{4 PV} = \frac{n_{1} RT}{n_{2} RT}$
$\frac{1}{4} = \frac{n_{1}}{n_{2}}$
$∴ n_{2} = 4 n_{1}$
$\frac{n_{2}}{n_{1}} = \frac{4}{1}$
$n_{2} : n_{1} = 4 : 1$

COMEDK 2019

Kinetic Theory of Gases

138996 An ideal gas at temperature $T$ , pressure $p$ occupies a volume $V$ . If its temperature is halved and pressure doubled, what is its new volume?

1

\frac{V}{4}

2

\frac{V}{2}

3

V

4

2 V

Explanation:

A According to combined gas law equation.
$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$
Where $P_{1}, V_{1}, T_{1}$ are the pressure, volume and temperature of the gas in initial state And
$P_{2}, V_{2}, T_{2}$ are the pressure volume and temperature of gas in final state.
$V_{2} = \frac{P_{1} V_{1} T_{2}}{T_{1} P_{2}}$
Given, initial temperature $(T_{1}) = T$ , Pressure $(P_{1}) = P$ and volume $(V_{1}) = V$
Final temperature $(T_{2}) = \frac{T}{2}$
Final pressure $(P_{2}) = 2 P$
Then, $V_{2} =$ ?
$V_{2} = \frac{P \times V \times T}{T \times 2 P \times 2}$
$V_{2} = \frac{V}{4}$

Shift-1]

Kinetic Theory of Gases

138998 A cyclic process on 1 mole of an ideal gas is shown in the figure. The temperatures of the gas at points $A$ and $B$ are respectively $200 K$ and $400 K$ . A total of $200 J$ of heat is taken out from the system in the process. What is the work done by the gas in the process $B \to C$ ? (Let $R = 8.3 J K^{- 1} {mol}^{- 1}$ )

1

- 1660 J

2

1660 J

3

- 1860 J

4

+ 1860 J

Explanation:

C Given, $n = 1$ mole, $Δ Q = - 200 J, R = 8.3 J K^{- 1} {mol}^{- 1}$

In processes $C \to A$ volume is constant therefore work done zero.
In the cyclic processes change in internal energy will be zero then $dU = 0$
According to first law of thermodynamic -
$dQ = dU + dW$
$dQ = 0 + dW$
$dQ = W_{AB} + W_{BC} + W_{CA}$
$- 200 J = W_{AB} + W_{BC} + 0$
Then, $- 200 J = P (V_{B} - V_{A}) + W_{BC}$
$- 200 J = nR (T_{B} - T_{A}) + W_{BC}$
$- 200 J = 1 \times 8.3 (400 - 200) + W_{BC}$
$- 200 J = 8.3 \times 200 + W_{BC}$
$- 200 J = 1660.0 + W_{BC}$
Then $W_{BC} = - 200 J - 1660 J = - 1860 J$

Shift-I]

Kinetic Theory of Gases

138999 Molecular weight of oxygen is 32 amu. At STP, volume of $1 gm$ of oxygen is $700 {cm}^{3}$ . What is the value of gas constant $R$ ?

1

18.23 J {Mol}^{- 1} K^{- 1}

2

27.73 J {mol}^{- 1} K^{- 1}

3

37.97 J {mol}^{- 1} K^{- 1}

4

8.31 J {mol}^{- 1} K^{- 1}

Explanation:

D Given, molecular weight of $O_{2} = 32 amu$ At STP, $P = 1 atm = 1.013 \times 10^{5}, T = 273 K$
$∴ 1 gm of O_{2} - 700 {cm}^{3} (given)$
$32 gm of O_{2} - 700 \times 32 {cm}^{3}$
$700 \times 32 \times 10^{- 6} m^{3}$
For 1 mole of an ideal gas equation-
$PV = RT$
$R = \frac{PV}{T} = \frac{1.013 \times 10^{5} \times 700 \times 32 \times 10^{- 6}}{273}$
$R = 8.31 {mol}^{- 1} K^{- 1}$

J and K-CET-2019

Kinetic Theory of Gases

139000 A thermally insulated piston divides a container into a two compartments. The volume, temperature and pressure in the right compartment are $2 V, T$ and $2 P$ , while in the left compartment the corresponding parameters are $V, T$ and $P$ . Then the ratio of number of molecules in the right compartment to that in the left compartment is

1

1 : 1

2

2 : 1

3

4 : 1

4

1 : 4

Explanation:

C According to Ideal gas equation $PV = nRT$
For left compartment,
$PV = \frac{n_{1}}{N_{A}} RT$
(Where, $N_{A} =$ Avogadro No.)
For right compartment
$(2 P) (2 V) = \frac{n_{2}}{N_{A}} RT$
Dividing equation (i) by (ii), we get
$\frac{PV}{4 PV} = \frac{n_{1} RT}{n_{2} RT}$
$\frac{1}{4} = \frac{n_{1}}{n_{2}}$
$∴ n_{2} = 4 n_{1}$
$\frac{n_{2}}{n_{1}} = \frac{4}{1}$
$n_{2} : n_{1} = 4 : 1$

COMEDK 2019

Kinetic Theory of Gases

138996 An ideal gas at temperature $T$ , pressure $p$ occupies a volume $V$ . If its temperature is halved and pressure doubled, what is its new volume?

1

\frac{V}{4}

2

\frac{V}{2}

3

V

4

2 V

Explanation:

A According to combined gas law equation.
$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$
Where $P_{1}, V_{1}, T_{1}$ are the pressure, volume and temperature of the gas in initial state And
$P_{2}, V_{2}, T_{2}$ are the pressure volume and temperature of gas in final state.
$V_{2} = \frac{P_{1} V_{1} T_{2}}{T_{1} P_{2}}$
Given, initial temperature $(T_{1}) = T$ , Pressure $(P_{1}) = P$ and volume $(V_{1}) = V$
Final temperature $(T_{2}) = \frac{T}{2}$
Final pressure $(P_{2}) = 2 P$
Then, $V_{2} =$ ?
$V_{2} = \frac{P \times V \times T}{T \times 2 P \times 2}$
$V_{2} = \frac{V}{4}$

Shift-1]

Kinetic Theory of Gases

138998 A cyclic process on 1 mole of an ideal gas is shown in the figure. The temperatures of the gas at points $A$ and $B$ are respectively $200 K$ and $400 K$ . A total of $200 J$ of heat is taken out from the system in the process. What is the work done by the gas in the process $B \to C$ ? (Let $R = 8.3 J K^{- 1} {mol}^{- 1}$ )

1

- 1660 J

2

1660 J

3

- 1860 J

4

+ 1860 J

Explanation:

C Given, $n = 1$ mole, $Δ Q = - 200 J, R = 8.3 J K^{- 1} {mol}^{- 1}$

In processes $C \to A$ volume is constant therefore work done zero.
In the cyclic processes change in internal energy will be zero then $dU = 0$
According to first law of thermodynamic -
$dQ = dU + dW$
$dQ = 0 + dW$
$dQ = W_{AB} + W_{BC} + W_{CA}$
$- 200 J = W_{AB} + W_{BC} + 0$
Then, $- 200 J = P (V_{B} - V_{A}) + W_{BC}$
$- 200 J = nR (T_{B} - T_{A}) + W_{BC}$
$- 200 J = 1 \times 8.3 (400 - 200) + W_{BC}$
$- 200 J = 8.3 \times 200 + W_{BC}$
$- 200 J = 1660.0 + W_{BC}$
Then $W_{BC} = - 200 J - 1660 J = - 1860 J$

Shift-I]

Kinetic Theory of Gases

138999 Molecular weight of oxygen is 32 amu. At STP, volume of $1 gm$ of oxygen is $700 {cm}^{3}$ . What is the value of gas constant $R$ ?

1

18.23 J {Mol}^{- 1} K^{- 1}

2

27.73 J {mol}^{- 1} K^{- 1}

3

37.97 J {mol}^{- 1} K^{- 1}

4

8.31 J {mol}^{- 1} K^{- 1}

Explanation:

D Given, molecular weight of $O_{2} = 32 amu$ At STP, $P = 1 atm = 1.013 \times 10^{5}, T = 273 K$
$∴ 1 gm of O_{2} - 700 {cm}^{3} (given)$
$32 gm of O_{2} - 700 \times 32 {cm}^{3}$
$700 \times 32 \times 10^{- 6} m^{3}$
For 1 mole of an ideal gas equation-
$PV = RT$
$R = \frac{PV}{T} = \frac{1.013 \times 10^{5} \times 700 \times 32 \times 10^{- 6}}{273}$
$R = 8.31 {mol}^{- 1} K^{- 1}$

J and K-CET-2019

Kinetic Theory of Gases

139000 A thermally insulated piston divides a container into a two compartments. The volume, temperature and pressure in the right compartment are $2 V, T$ and $2 P$ , while in the left compartment the corresponding parameters are $V, T$ and $P$ . Then the ratio of number of molecules in the right compartment to that in the left compartment is

1

1 : 1

2

2 : 1

3

4 : 1

4

1 : 4

Explanation:

C According to Ideal gas equation $PV = nRT$
For left compartment,
$PV = \frac{n_{1}}{N_{A}} RT$
(Where, $N_{A} =$ Avogadro No.)
For right compartment
$(2 P) (2 V) = \frac{n_{2}}{N_{A}} RT$
Dividing equation (i) by (ii), we get
$\frac{PV}{4 PV} = \frac{n_{1} RT}{n_{2} RT}$
$\frac{1}{4} = \frac{n_{1}}{n_{2}}$
$∴ n_{2} = 4 n_{1}$
$\frac{n_{2}}{n_{1}} = \frac{4}{1}$
$n_{2} : n_{1} = 4 : 1$

COMEDK 2019

Kinetic Theory of Gases

138996 An ideal gas at temperature $T$ , pressure $p$ occupies a volume $V$ . If its temperature is halved and pressure doubled, what is its new volume?

1

\frac{V}{4}

2

\frac{V}{2}

3

V

4

2 V

Explanation:

A According to combined gas law equation.
$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$
Where $P_{1}, V_{1}, T_{1}$ are the pressure, volume and temperature of the gas in initial state And
$P_{2}, V_{2}, T_{2}$ are the pressure volume and temperature of gas in final state.
$V_{2} = \frac{P_{1} V_{1} T_{2}}{T_{1} P_{2}}$
Given, initial temperature $(T_{1}) = T$ , Pressure $(P_{1}) = P$ and volume $(V_{1}) = V$
Final temperature $(T_{2}) = \frac{T}{2}$
Final pressure $(P_{2}) = 2 P$
Then, $V_{2} =$ ?
$V_{2} = \frac{P \times V \times T}{T \times 2 P \times 2}$
$V_{2} = \frac{V}{4}$

Shift-1]

Kinetic Theory of Gases

138998 A cyclic process on 1 mole of an ideal gas is shown in the figure. The temperatures of the gas at points $A$ and $B$ are respectively $200 K$ and $400 K$ . A total of $200 J$ of heat is taken out from the system in the process. What is the work done by the gas in the process $B \to C$ ? (Let $R = 8.3 J K^{- 1} {mol}^{- 1}$ )

1

- 1660 J

2

1660 J

3

- 1860 J

4

+ 1860 J

Explanation:

C Given, $n = 1$ mole, $Δ Q = - 200 J, R = 8.3 J K^{- 1} {mol}^{- 1}$

In processes $C \to A$ volume is constant therefore work done zero.
In the cyclic processes change in internal energy will be zero then $dU = 0$
According to first law of thermodynamic -
$dQ = dU + dW$
$dQ = 0 + dW$
$dQ = W_{AB} + W_{BC} + W_{CA}$
$- 200 J = W_{AB} + W_{BC} + 0$
Then, $- 200 J = P (V_{B} - V_{A}) + W_{BC}$
$- 200 J = nR (T_{B} - T_{A}) + W_{BC}$
$- 200 J = 1 \times 8.3 (400 - 200) + W_{BC}$
$- 200 J = 8.3 \times 200 + W_{BC}$
$- 200 J = 1660.0 + W_{BC}$
Then $W_{BC} = - 200 J - 1660 J = - 1860 J$

Shift-I]

Kinetic Theory of Gases

138999 Molecular weight of oxygen is 32 amu. At STP, volume of $1 gm$ of oxygen is $700 {cm}^{3}$ . What is the value of gas constant $R$ ?

1

18.23 J {Mol}^{- 1} K^{- 1}

2

27.73 J {mol}^{- 1} K^{- 1}

3

37.97 J {mol}^{- 1} K^{- 1}

4

8.31 J {mol}^{- 1} K^{- 1}

Explanation:

D Given, molecular weight of $O_{2} = 32 amu$ At STP, $P = 1 atm = 1.013 \times 10^{5}, T = 273 K$
$∴ 1 gm of O_{2} - 700 {cm}^{3} (given)$
$32 gm of O_{2} - 700 \times 32 {cm}^{3}$
$700 \times 32 \times 10^{- 6} m^{3}$
For 1 mole of an ideal gas equation-
$PV = RT$
$R = \frac{PV}{T} = \frac{1.013 \times 10^{5} \times 700 \times 32 \times 10^{- 6}}{273}$
$R = 8.31 {mol}^{- 1} K^{- 1}$

J and K-CET-2019

Kinetic Theory of Gases

139000 A thermally insulated piston divides a container into a two compartments. The volume, temperature and pressure in the right compartment are $2 V, T$ and $2 P$ , while in the left compartment the corresponding parameters are $V, T$ and $P$ . Then the ratio of number of molecules in the right compartment to that in the left compartment is

1

1 : 1

2

2 : 1

3

4 : 1

4

1 : 4

Explanation:

C According to Ideal gas equation $PV = nRT$
For left compartment,
$PV = \frac{n_{1}}{N_{A}} RT$
(Where, $N_{A} =$ Avogadro No.)
For right compartment
$(2 P) (2 V) = \frac{n_{2}}{N_{A}} RT$
Dividing equation (i) by (ii), we get
$\frac{PV}{4 PV} = \frac{n_{1} RT}{n_{2} RT}$
$\frac{1}{4} = \frac{n_{1}}{n_{2}}$
$∴ n_{2} = 4 n_{1}$
$\frac{n_{2}}{n_{1}} = \frac{4}{1}$
$n_{2} : n_{1} = 4 : 1$

COMEDK 2019