QBManager

Thermodynamics

148492 A mixture of two non-reactive ideal gases is enclosed in a vessel consisting of one mole of a monoatomic gas ' $A$ ' and ' $n$ ' moles of diatomic gas ' $B$ ' at a temperature ' $T$ '. If the adiabatic constant of the gaseous mixture is $\frac{13}{9}$ , then the value of ' $n$ ' is.

1 5

2 2

3 4

4 3

Explanation:

D | monoatomic | diatomic |
| :--- | :--- |
| $n_{1} = 1$ | $n_{2} = n$ |
| $C_{v_{1}} = \frac{3}{2} R$ | $C_{v_{2}} = \frac{5}{2} R$ |
${(C_{v})}_{mix} = \frac{n_{1} C_{V_{1}} + n_{2} C_{V_{2}}}{n_{1} + n_{2}}$
${(C_{v})}_{mix} = \frac{\frac{3}{2} R + n \frac{5}{2} R}{1 + n}$
$= \frac{R}{2} \frac{(3 + 5 n)}{(1 + n)}$
${(C_{P})}_{mix} = {(C_{v})}_{mix} + R$
$= \frac{R}{2} (\frac{3 + 5 n}{1 + n}) + R$
${(C_{P})}_{mix} = \frac{R}{2} (\frac{5 + 7 n}{1 + n})$
$(γ_{mix}) = \frac{{(C_{P})}_{mix}}{{(C_{v})}_{mix}}$
$= \frac{\frac{R}{2} (\frac{5 + 7 n}{1 + n})}{\frac{R}{2} (\frac{3 + 5 n}{1 + n})}$
$\frac{13}{9} = \frac{5 + 7 n}{3 + 5 n}$
$39 + 65 n = 45 + 63 n$
$65 n - 63 n = 6$
$2 n = 6$
$n = 3$

AP EAMCET-25.04.2017

Thermodynamics

148493 5.6 litre of helium gas at STP is adiabatically compressed to 0.7 litre. If the initial temperature of the gas is $T K$ , work done in the process is ( $R$ is universal gas constant in SI units)

1

\frac{9}{8} RT

2

- (\frac{9}{8} RT)

3

- (\frac{4}{3} RT)

4

\frac{3}{4} RT

Explanation:

B Given that,
Initial volume $= 5.6 l$
Final volume $= 0.7 l$
Initial temperature $= T K$
${TV}^{γ - 1} = c$
$T_{1} {(V_{1})}^{γ - 1} = T_{2} {(V_{2})}^{γ - 1}$
$T_{1} (5.6)^{2 / 3} = T_{2} (0.7)^{2 / 3}$
$γ = \frac{5}{3}$ for monoatomic
$T_{2} = (8)^{2 / 3} T_{1}$
$T_{2} = {(2^{3})}^{2 / 3} T_{1}$
$T_{2} = 4 T_{1}$
Work done, $W = \frac{nR Δ T}{(γ - 1)}$
$= \frac{\frac{1}{4} R \cdot (T - 4 T)}{\frac{5}{3} - 1}$
$= \frac{- 3 R T / 4}{2 / 3}$
$= - \frac{9 R T}{8}$

AP EAMCET-25.04.2017

Thermodynamics

148494 Starting with the same initial conditions, an ideal gas expands from volume $V_{1}$ to $V_{2}$ in three different ways. The work done by the gas is $W_{1}$ if the process is purely isothermal, $W_{2}$ if purely isobaric and $W_{3}$ if purely adiabatic. The

1

W_{1} > W_{2} > W_{3}

2

W_{2} > W_{1} > W_{3}

3

W_{2} > W_{3} > W_{1}

4

W_{1} > W_{3} > W_{2}

Explanation:

B Work done area under the $P - V$ diagram
Work done for isobaric $= P (V_{2} - V_{1})$
Work done for isothermal $= P_{1} V_{1} \log \frac{V_{2}}{V_{1}}$
Work done for adiabatic $= \frac{P_{2} V_{2} - P_{1} V_{1}}{γ - 1}$

Since, area under the curve is maximum for isobaric process in expansion.

AMU-2008

Thermodynamics

148495 An ideal gas is compressed to half its initial volume by means of several process. Which of the process results in the maximum work done on the gas?

1 Adiabatic

2 Isobaric

3 Isochoric

4 Isothermal

Explanation:

A
Work done on the gas $=$ area under $P - V$ curve.
As area under the $P - V$ curve is maximum for adiabatic process, so work done on the gas is maximum for adiabatic process.
$(Slope)_{adia} > (slope)_{isothermal}$
$(Area)_{adia} > (Area)_{iso} > (Area)_{isobaric}$

AIPMT-2015

Thermodynamics

148498 In which of the following processes, heat is neither absorbed nor released by a system?

1 Adiabatic

2 Isobaric

3 Isochoric

4 Isothermal

Explanation:

A Adiabatic process:
In adiabatic process exchange of heat $Δ Q = 0$
An adiabatic process is a reversible process with constant entropy for an ideal gas.
Isothermal process:
It is a process in which temperature remains constant.
Isobaric process - pressure is constant
Isochoric process - Volume is constant

NEET National-2019

Thermodynamics

148492 A mixture of two non-reactive ideal gases is enclosed in a vessel consisting of one mole of a monoatomic gas ' $A$ ' and ' $n$ ' moles of diatomic gas ' $B$ ' at a temperature ' $T$ '. If the adiabatic constant of the gaseous mixture is $\frac{13}{9}$ , then the value of ' $n$ ' is.

1 5

2 2

3 4

4 3

Explanation:

D | monoatomic | diatomic |
| :--- | :--- |
| $n_{1} = 1$ | $n_{2} = n$ |
| $C_{v_{1}} = \frac{3}{2} R$ | $C_{v_{2}} = \frac{5}{2} R$ |
${(C_{v})}_{mix} = \frac{n_{1} C_{V_{1}} + n_{2} C_{V_{2}}}{n_{1} + n_{2}}$
${(C_{v})}_{mix} = \frac{\frac{3}{2} R + n \frac{5}{2} R}{1 + n}$
$= \frac{R}{2} \frac{(3 + 5 n)}{(1 + n)}$
${(C_{P})}_{mix} = {(C_{v})}_{mix} + R$
$= \frac{R}{2} (\frac{3 + 5 n}{1 + n}) + R$
${(C_{P})}_{mix} = \frac{R}{2} (\frac{5 + 7 n}{1 + n})$
$(γ_{mix}) = \frac{{(C_{P})}_{mix}}{{(C_{v})}_{mix}}$
$= \frac{\frac{R}{2} (\frac{5 + 7 n}{1 + n})}{\frac{R}{2} (\frac{3 + 5 n}{1 + n})}$
$\frac{13}{9} = \frac{5 + 7 n}{3 + 5 n}$
$39 + 65 n = 45 + 63 n$
$65 n - 63 n = 6$
$2 n = 6$
$n = 3$

AP EAMCET-25.04.2017

Thermodynamics

148493 5.6 litre of helium gas at STP is adiabatically compressed to 0.7 litre. If the initial temperature of the gas is $T K$ , work done in the process is ( $R$ is universal gas constant in SI units)

1

\frac{9}{8} RT

2

- (\frac{9}{8} RT)

3

- (\frac{4}{3} RT)

4

\frac{3}{4} RT

Explanation:

B Given that,
Initial volume $= 5.6 l$
Final volume $= 0.7 l$
Initial temperature $= T K$
${TV}^{γ - 1} = c$
$T_{1} {(V_{1})}^{γ - 1} = T_{2} {(V_{2})}^{γ - 1}$
$T_{1} (5.6)^{2 / 3} = T_{2} (0.7)^{2 / 3}$
$γ = \frac{5}{3}$ for monoatomic
$T_{2} = (8)^{2 / 3} T_{1}$
$T_{2} = {(2^{3})}^{2 / 3} T_{1}$
$T_{2} = 4 T_{1}$
Work done, $W = \frac{nR Δ T}{(γ - 1)}$
$= \frac{\frac{1}{4} R \cdot (T - 4 T)}{\frac{5}{3} - 1}$
$= \frac{- 3 R T / 4}{2 / 3}$
$= - \frac{9 R T}{8}$

AP EAMCET-25.04.2017

Thermodynamics

148494 Starting with the same initial conditions, an ideal gas expands from volume $V_{1}$ to $V_{2}$ in three different ways. The work done by the gas is $W_{1}$ if the process is purely isothermal, $W_{2}$ if purely isobaric and $W_{3}$ if purely adiabatic. The

1

W_{1} > W_{2} > W_{3}

2

W_{2} > W_{1} > W_{3}

3

W_{2} > W_{3} > W_{1}

4

W_{1} > W_{3} > W_{2}

Explanation:

B Work done area under the $P - V$ diagram
Work done for isobaric $= P (V_{2} - V_{1})$
Work done for isothermal $= P_{1} V_{1} \log \frac{V_{2}}{V_{1}}$
Work done for adiabatic $= \frac{P_{2} V_{2} - P_{1} V_{1}}{γ - 1}$

Since, area under the curve is maximum for isobaric process in expansion.

AMU-2008

Thermodynamics

148495 An ideal gas is compressed to half its initial volume by means of several process. Which of the process results in the maximum work done on the gas?

1 Adiabatic

2 Isobaric

3 Isochoric

4 Isothermal

Explanation:

A
Work done on the gas $=$ area under $P - V$ curve.
As area under the $P - V$ curve is maximum for adiabatic process, so work done on the gas is maximum for adiabatic process.
$(Slope)_{adia} > (slope)_{isothermal}$
$(Area)_{adia} > (Area)_{iso} > (Area)_{isobaric}$

AIPMT-2015

Thermodynamics

148498 In which of the following processes, heat is neither absorbed nor released by a system?

1 Adiabatic

2 Isobaric

3 Isochoric

4 Isothermal

Explanation:

A Adiabatic process:
In adiabatic process exchange of heat $Δ Q = 0$
An adiabatic process is a reversible process with constant entropy for an ideal gas.
Isothermal process:
It is a process in which temperature remains constant.
Isobaric process - pressure is constant
Isochoric process - Volume is constant

NEET National-2019

Thermodynamics

148492 A mixture of two non-reactive ideal gases is enclosed in a vessel consisting of one mole of a monoatomic gas ' $A$ ' and ' $n$ ' moles of diatomic gas ' $B$ ' at a temperature ' $T$ '. If the adiabatic constant of the gaseous mixture is $\frac{13}{9}$ , then the value of ' $n$ ' is.

1 5

2 2

3 4

4 3

Explanation:

D | monoatomic | diatomic |
| :--- | :--- |
| $n_{1} = 1$ | $n_{2} = n$ |
| $C_{v_{1}} = \frac{3}{2} R$ | $C_{v_{2}} = \frac{5}{2} R$ |
${(C_{v})}_{mix} = \frac{n_{1} C_{V_{1}} + n_{2} C_{V_{2}}}{n_{1} + n_{2}}$
${(C_{v})}_{mix} = \frac{\frac{3}{2} R + n \frac{5}{2} R}{1 + n}$
$= \frac{R}{2} \frac{(3 + 5 n)}{(1 + n)}$
${(C_{P})}_{mix} = {(C_{v})}_{mix} + R$
$= \frac{R}{2} (\frac{3 + 5 n}{1 + n}) + R$
${(C_{P})}_{mix} = \frac{R}{2} (\frac{5 + 7 n}{1 + n})$
$(γ_{mix}) = \frac{{(C_{P})}_{mix}}{{(C_{v})}_{mix}}$
$= \frac{\frac{R}{2} (\frac{5 + 7 n}{1 + n})}{\frac{R}{2} (\frac{3 + 5 n}{1 + n})}$
$\frac{13}{9} = \frac{5 + 7 n}{3 + 5 n}$
$39 + 65 n = 45 + 63 n$
$65 n - 63 n = 6$
$2 n = 6$
$n = 3$

AP EAMCET-25.04.2017

Thermodynamics

148493 5.6 litre of helium gas at STP is adiabatically compressed to 0.7 litre. If the initial temperature of the gas is $T K$ , work done in the process is ( $R$ is universal gas constant in SI units)

1

\frac{9}{8} RT

2

- (\frac{9}{8} RT)

3

- (\frac{4}{3} RT)

4

\frac{3}{4} RT

Explanation:

B Given that,
Initial volume $= 5.6 l$
Final volume $= 0.7 l$
Initial temperature $= T K$
${TV}^{γ - 1} = c$
$T_{1} {(V_{1})}^{γ - 1} = T_{2} {(V_{2})}^{γ - 1}$
$T_{1} (5.6)^{2 / 3} = T_{2} (0.7)^{2 / 3}$
$γ = \frac{5}{3}$ for monoatomic
$T_{2} = (8)^{2 / 3} T_{1}$
$T_{2} = {(2^{3})}^{2 / 3} T_{1}$
$T_{2} = 4 T_{1}$
Work done, $W = \frac{nR Δ T}{(γ - 1)}$
$= \frac{\frac{1}{4} R \cdot (T - 4 T)}{\frac{5}{3} - 1}$
$= \frac{- 3 R T / 4}{2 / 3}$
$= - \frac{9 R T}{8}$

AP EAMCET-25.04.2017

Thermodynamics

148494 Starting with the same initial conditions, an ideal gas expands from volume $V_{1}$ to $V_{2}$ in three different ways. The work done by the gas is $W_{1}$ if the process is purely isothermal, $W_{2}$ if purely isobaric and $W_{3}$ if purely adiabatic. The

1

W_{1} > W_{2} > W_{3}

2

W_{2} > W_{1} > W_{3}

3

W_{2} > W_{3} > W_{1}

4

W_{1} > W_{3} > W_{2}

Explanation:

B Work done area under the $P - V$ diagram
Work done for isobaric $= P (V_{2} - V_{1})$
Work done for isothermal $= P_{1} V_{1} \log \frac{V_{2}}{V_{1}}$
Work done for adiabatic $= \frac{P_{2} V_{2} - P_{1} V_{1}}{γ - 1}$

Since, area under the curve is maximum for isobaric process in expansion.

AMU-2008

Thermodynamics

148495 An ideal gas is compressed to half its initial volume by means of several process. Which of the process results in the maximum work done on the gas?

1 Adiabatic

2 Isobaric

3 Isochoric

4 Isothermal

Explanation:

A
Work done on the gas $=$ area under $P - V$ curve.
As area under the $P - V$ curve is maximum for adiabatic process, so work done on the gas is maximum for adiabatic process.
$(Slope)_{adia} > (slope)_{isothermal}$
$(Area)_{adia} > (Area)_{iso} > (Area)_{isobaric}$

AIPMT-2015

Thermodynamics

148498 In which of the following processes, heat is neither absorbed nor released by a system?

1 Adiabatic

2 Isobaric

3 Isochoric

4 Isothermal

Explanation:

A Adiabatic process:
In adiabatic process exchange of heat $Δ Q = 0$
An adiabatic process is a reversible process with constant entropy for an ideal gas.
Isothermal process:
It is a process in which temperature remains constant.
Isobaric process - pressure is constant
Isochoric process - Volume is constant

NEET National-2019

Thermodynamics

148492 A mixture of two non-reactive ideal gases is enclosed in a vessel consisting of one mole of a monoatomic gas ' $A$ ' and ' $n$ ' moles of diatomic gas ' $B$ ' at a temperature ' $T$ '. If the adiabatic constant of the gaseous mixture is $\frac{13}{9}$ , then the value of ' $n$ ' is.

1 5

2 2

3 4

4 3

Explanation:

D | monoatomic | diatomic |
| :--- | :--- |
| $n_{1} = 1$ | $n_{2} = n$ |
| $C_{v_{1}} = \frac{3}{2} R$ | $C_{v_{2}} = \frac{5}{2} R$ |
${(C_{v})}_{mix} = \frac{n_{1} C_{V_{1}} + n_{2} C_{V_{2}}}{n_{1} + n_{2}}$
${(C_{v})}_{mix} = \frac{\frac{3}{2} R + n \frac{5}{2} R}{1 + n}$
$= \frac{R}{2} \frac{(3 + 5 n)}{(1 + n)}$
${(C_{P})}_{mix} = {(C_{v})}_{mix} + R$
$= \frac{R}{2} (\frac{3 + 5 n}{1 + n}) + R$
${(C_{P})}_{mix} = \frac{R}{2} (\frac{5 + 7 n}{1 + n})$
$(γ_{mix}) = \frac{{(C_{P})}_{mix}}{{(C_{v})}_{mix}}$
$= \frac{\frac{R}{2} (\frac{5 + 7 n}{1 + n})}{\frac{R}{2} (\frac{3 + 5 n}{1 + n})}$
$\frac{13}{9} = \frac{5 + 7 n}{3 + 5 n}$
$39 + 65 n = 45 + 63 n$
$65 n - 63 n = 6$
$2 n = 6$
$n = 3$

AP EAMCET-25.04.2017

Thermodynamics

148493 5.6 litre of helium gas at STP is adiabatically compressed to 0.7 litre. If the initial temperature of the gas is $T K$ , work done in the process is ( $R$ is universal gas constant in SI units)

1

\frac{9}{8} RT

2

- (\frac{9}{8} RT)

3

- (\frac{4}{3} RT)

4

\frac{3}{4} RT

Explanation:

B Given that,
Initial volume $= 5.6 l$
Final volume $= 0.7 l$
Initial temperature $= T K$
${TV}^{γ - 1} = c$
$T_{1} {(V_{1})}^{γ - 1} = T_{2} {(V_{2})}^{γ - 1}$
$T_{1} (5.6)^{2 / 3} = T_{2} (0.7)^{2 / 3}$
$γ = \frac{5}{3}$ for monoatomic
$T_{2} = (8)^{2 / 3} T_{1}$
$T_{2} = {(2^{3})}^{2 / 3} T_{1}$
$T_{2} = 4 T_{1}$
Work done, $W = \frac{nR Δ T}{(γ - 1)}$
$= \frac{\frac{1}{4} R \cdot (T - 4 T)}{\frac{5}{3} - 1}$
$= \frac{- 3 R T / 4}{2 / 3}$
$= - \frac{9 R T}{8}$

AP EAMCET-25.04.2017

Thermodynamics

148494 Starting with the same initial conditions, an ideal gas expands from volume $V_{1}$ to $V_{2}$ in three different ways. The work done by the gas is $W_{1}$ if the process is purely isothermal, $W_{2}$ if purely isobaric and $W_{3}$ if purely adiabatic. The

1

W_{1} > W_{2} > W_{3}

2

W_{2} > W_{1} > W_{3}

3

W_{2} > W_{3} > W_{1}

4

W_{1} > W_{3} > W_{2}

Explanation:

B Work done area under the $P - V$ diagram
Work done for isobaric $= P (V_{2} - V_{1})$
Work done for isothermal $= P_{1} V_{1} \log \frac{V_{2}}{V_{1}}$
Work done for adiabatic $= \frac{P_{2} V_{2} - P_{1} V_{1}}{γ - 1}$

Since, area under the curve is maximum for isobaric process in expansion.

AMU-2008

Thermodynamics

148495 An ideal gas is compressed to half its initial volume by means of several process. Which of the process results in the maximum work done on the gas?

1 Adiabatic

2 Isobaric

3 Isochoric

4 Isothermal

Explanation:

A
Work done on the gas $=$ area under $P - V$ curve.
As area under the $P - V$ curve is maximum for adiabatic process, so work done on the gas is maximum for adiabatic process.
$(Slope)_{adia} > (slope)_{isothermal}$
$(Area)_{adia} > (Area)_{iso} > (Area)_{isobaric}$

AIPMT-2015

Thermodynamics

148498 In which of the following processes, heat is neither absorbed nor released by a system?

1 Adiabatic

2 Isobaric

3 Isochoric

4 Isothermal

Explanation:

A Adiabatic process:
In adiabatic process exchange of heat $Δ Q = 0$
An adiabatic process is a reversible process with constant entropy for an ideal gas.
Isothermal process:
It is a process in which temperature remains constant.
Isobaric process - pressure is constant
Isochoric process - Volume is constant

NEET National-2019

Thermodynamics

148492 A mixture of two non-reactive ideal gases is enclosed in a vessel consisting of one mole of a monoatomic gas ' $A$ ' and ' $n$ ' moles of diatomic gas ' $B$ ' at a temperature ' $T$ '. If the adiabatic constant of the gaseous mixture is $\frac{13}{9}$ , then the value of ' $n$ ' is.

1 5

2 2

3 4

4 3

Explanation:

D | monoatomic | diatomic |
| :--- | :--- |
| $n_{1} = 1$ | $n_{2} = n$ |
| $C_{v_{1}} = \frac{3}{2} R$ | $C_{v_{2}} = \frac{5}{2} R$ |
${(C_{v})}_{mix} = \frac{n_{1} C_{V_{1}} + n_{2} C_{V_{2}}}{n_{1} + n_{2}}$
${(C_{v})}_{mix} = \frac{\frac{3}{2} R + n \frac{5}{2} R}{1 + n}$
$= \frac{R}{2} \frac{(3 + 5 n)}{(1 + n)}$
${(C_{P})}_{mix} = {(C_{v})}_{mix} + R$
$= \frac{R}{2} (\frac{3 + 5 n}{1 + n}) + R$
${(C_{P})}_{mix} = \frac{R}{2} (\frac{5 + 7 n}{1 + n})$
$(γ_{mix}) = \frac{{(C_{P})}_{mix}}{{(C_{v})}_{mix}}$
$= \frac{\frac{R}{2} (\frac{5 + 7 n}{1 + n})}{\frac{R}{2} (\frac{3 + 5 n}{1 + n})}$
$\frac{13}{9} = \frac{5 + 7 n}{3 + 5 n}$
$39 + 65 n = 45 + 63 n$
$65 n - 63 n = 6$
$2 n = 6$
$n = 3$

AP EAMCET-25.04.2017

Thermodynamics

148493 5.6 litre of helium gas at STP is adiabatically compressed to 0.7 litre. If the initial temperature of the gas is $T K$ , work done in the process is ( $R$ is universal gas constant in SI units)

1

\frac{9}{8} RT

2

- (\frac{9}{8} RT)

3

- (\frac{4}{3} RT)

4

\frac{3}{4} RT

Explanation:

B Given that,
Initial volume $= 5.6 l$
Final volume $= 0.7 l$
Initial temperature $= T K$
${TV}^{γ - 1} = c$
$T_{1} {(V_{1})}^{γ - 1} = T_{2} {(V_{2})}^{γ - 1}$
$T_{1} (5.6)^{2 / 3} = T_{2} (0.7)^{2 / 3}$
$γ = \frac{5}{3}$ for monoatomic
$T_{2} = (8)^{2 / 3} T_{1}$
$T_{2} = {(2^{3})}^{2 / 3} T_{1}$
$T_{2} = 4 T_{1}$
Work done, $W = \frac{nR Δ T}{(γ - 1)}$
$= \frac{\frac{1}{4} R \cdot (T - 4 T)}{\frac{5}{3} - 1}$
$= \frac{- 3 R T / 4}{2 / 3}$
$= - \frac{9 R T}{8}$

AP EAMCET-25.04.2017

Thermodynamics

148494 Starting with the same initial conditions, an ideal gas expands from volume $V_{1}$ to $V_{2}$ in three different ways. The work done by the gas is $W_{1}$ if the process is purely isothermal, $W_{2}$ if purely isobaric and $W_{3}$ if purely adiabatic. The

1

W_{1} > W_{2} > W_{3}

2

W_{2} > W_{1} > W_{3}

3

W_{2} > W_{3} > W_{1}

4

W_{1} > W_{3} > W_{2}

Explanation:

B Work done area under the $P - V$ diagram
Work done for isobaric $= P (V_{2} - V_{1})$
Work done for isothermal $= P_{1} V_{1} \log \frac{V_{2}}{V_{1}}$
Work done for adiabatic $= \frac{P_{2} V_{2} - P_{1} V_{1}}{γ - 1}$

Since, area under the curve is maximum for isobaric process in expansion.

AMU-2008

Thermodynamics

148495 An ideal gas is compressed to half its initial volume by means of several process. Which of the process results in the maximum work done on the gas?

1 Adiabatic

2 Isobaric

3 Isochoric

4 Isothermal

Explanation:

A
Work done on the gas $=$ area under $P - V$ curve.
As area under the $P - V$ curve is maximum for adiabatic process, so work done on the gas is maximum for adiabatic process.
$(Slope)_{adia} > (slope)_{isothermal}$
$(Area)_{adia} > (Area)_{iso} > (Area)_{isobaric}$

AIPMT-2015

Thermodynamics

148498 In which of the following processes, heat is neither absorbed nor released by a system?

1 Adiabatic

2 Isobaric

3 Isochoric

4 Isothermal

Explanation:

A Adiabatic process:
In adiabatic process exchange of heat $Δ Q = 0$
An adiabatic process is a reversible process with constant entropy for an ideal gas.
Isothermal process:
It is a process in which temperature remains constant.
Isobaric process - pressure is constant
Isochoric process - Volume is constant

NEET National-2019

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