QBManager

Thermodynamics

148436 A hypothetical gas expands adiabatically such that its volume changes from 08 litres to 27 liters. If the ratio of final pressure of the gas to initial pressure of the gas is $\frac{16}{81}$ . Then the ratio of $\frac{c_{p}}{c_{v}}$ will be.

1

\frac{3}{1}

2

\frac{3}{2}

3

\frac{1}{2}

4

\frac{4}{3}

Explanation:

D We know, $\frac{c_{p}}{c_{v}} = γ$
For an adiabatic process,
${PV}^{γ} = C$
$P_{1} V_{1}^{γ} = P_{2} V_{2}^{γ}$
$\frac{P_{1}}{P_{2}} = {(\frac{V_{2}}{V_{1}})}^{γ}$
Given, $\frac{P_{2}}{P_{1}} = \frac{16}{81}$
$\frac{81}{16} = {(\frac{27}{8})}^{γ}$
${(\frac{3}{2})}^{4} = {(\frac{3}{2})}^{3 γ}$
By comparing,
$4 = 3 γ$
$\frac{c_{p}}{c_{v}} = γ = \frac{4}{3}$

JEE Main-31.01.2023

Thermodynamics

148439 The ratio of the adiabatic to isothermal elasticities of a triatomic (non-linear) gas is

1

3 : 4

2

1 : 2

3

4 : 3

4

5 : 3

Explanation:

C From elasticity of adiabatic process
$E_{adi} = γ P$
From elasticity of isothermal process
$E_{iso} = P$
Ratio of the adiabatic to isothermal elasticity,
$\frac{E_{adi}}{E_{iso}} = \frac{γ P}{P} = γ = 1 + \frac{2}{f} [For triatomic, f = 6]$
$= 1 + \frac{2}{6} = \frac{4}{3}$

AP EAMCET (21.09.2020) Shift-II

Thermodynamics

148440 Five moles of Hydrogen gas initially at STP is compressed adiabatically so that its temperature becomes $673 K$ . The increase in internal energy of the gas is $(R = 8.3 J {mol}^{- 1} K^{- 1}, γ = 1.4$ for diatomic gas)

1

80.5 kJ

2

21.55 kJ

3

41.50 kJ

4

65.55 kJ

Explanation:

C Final temperature $= 673 K = T_{2}$ Initial temperature $= 273 K = T_{1}$ Number of moles of hydrogen, $n = 5$ Universal gas constant $(R) = 8.3 J {mol}^{- 1} K^{- 1}$ For diatomic gas $(γ) = 1.4$
Now, change in internal energy due to adiabatic process is
$∴ Δ U = {nc}_{v} Δ T$
$Δ U = 5 \times \frac{R}{γ - 1} (T_{2} - T_{1}) (\begin{matrix} ∵ Specific heat, C_{v} = \frac{R}{γ - 1} \end{matrix})$
$= 5 \times \frac{8.3}{1.4 - 1} (673 - 273)$
$= 5 \times \frac{8.3}{0.4} \times 400$
$Δ U = 41.5 kJ$
So, the increase in internal energy of the gas is
$Δ U = 41.5 kJ$

AP EAMCET (20.04.2019) Shift-II

Thermodynamics

148441 In an adiabatic process, the pressure is increased by $\frac{2}{3} %$ .If $γ = \frac{2}{3}$ , then the volume decreases by nearly

1

\frac{4}{9} %

2

\frac{2}{3} %

3

1 %

4

\frac{9}{4} %

Explanation:

C For the adiabatic process.
${PV}^{γ} = Constant$
${PV}^{γ} = C$
Taking $\log$ on both sides.
Differentiating,
$\log P + γ \log V = \log C$
$\frac{Δ P}{P} + γ \frac{Δ V}{V} = 0$
$\frac{Δ V}{V} = \frac{- 1}{γ} (\frac{Δ P}{P})$
$\frac{Δ V}{V} \times 100 = \frac{- 1}{γ} (\frac{Δ P}{P} \times 100)$
$= - \frac{3}{2} \times \frac{2}{3}$
$= - 1 %$
So, volume decrease by $1 %$ .

BITSAT-2020

Thermodynamics

148442 Which equation is valid for adiabatic process?

1

{TV}^{γ - 1} =

constant

2

{pV}^{γ - 1} =

constant

3

T^{γ} V^{γ - 1} =

constant

4

\frac{p^{γ - 1}}{T^{γ - 1}} =

constant

Explanation:

A For an adiabatic process-
${PV}^{γ} =$ Constant
For an ideal gas $\frac{P V}{T} = n \bar{R} = C$
$P = \frac{C T}{V}$
$P = \frac{T}{V}$
From equation (i) and (ii)-
$(\frac{T}{V}) V^{γ} = constant$
$T . V^{γ - 1} = constant$

CG PET- 2014

Thermodynamics

148436 A hypothetical gas expands adiabatically such that its volume changes from 08 litres to 27 liters. If the ratio of final pressure of the gas to initial pressure of the gas is $\frac{16}{81}$ . Then the ratio of $\frac{c_{p}}{c_{v}}$ will be.

1

\frac{3}{1}

2

\frac{3}{2}

3

\frac{1}{2}

4

\frac{4}{3}

Explanation:

D We know, $\frac{c_{p}}{c_{v}} = γ$
For an adiabatic process,
${PV}^{γ} = C$
$P_{1} V_{1}^{γ} = P_{2} V_{2}^{γ}$
$\frac{P_{1}}{P_{2}} = {(\frac{V_{2}}{V_{1}})}^{γ}$
Given, $\frac{P_{2}}{P_{1}} = \frac{16}{81}$
$\frac{81}{16} = {(\frac{27}{8})}^{γ}$
${(\frac{3}{2})}^{4} = {(\frac{3}{2})}^{3 γ}$
By comparing,
$4 = 3 γ$
$\frac{c_{p}}{c_{v}} = γ = \frac{4}{3}$

JEE Main-31.01.2023

Thermodynamics

148439 The ratio of the adiabatic to isothermal elasticities of a triatomic (non-linear) gas is

1

3 : 4

2

1 : 2

3

4 : 3

4

5 : 3

Explanation:

C From elasticity of adiabatic process
$E_{adi} = γ P$
From elasticity of isothermal process
$E_{iso} = P$
Ratio of the adiabatic to isothermal elasticity,
$\frac{E_{adi}}{E_{iso}} = \frac{γ P}{P} = γ = 1 + \frac{2}{f} [For triatomic, f = 6]$
$= 1 + \frac{2}{6} = \frac{4}{3}$

AP EAMCET (21.09.2020) Shift-II

Thermodynamics

148440 Five moles of Hydrogen gas initially at STP is compressed adiabatically so that its temperature becomes $673 K$ . The increase in internal energy of the gas is $(R = 8.3 J {mol}^{- 1} K^{- 1}, γ = 1.4$ for diatomic gas)

1

80.5 kJ

2

21.55 kJ

3

41.50 kJ

4

65.55 kJ

Explanation:

C Final temperature $= 673 K = T_{2}$ Initial temperature $= 273 K = T_{1}$ Number of moles of hydrogen, $n = 5$ Universal gas constant $(R) = 8.3 J {mol}^{- 1} K^{- 1}$ For diatomic gas $(γ) = 1.4$
Now, change in internal energy due to adiabatic process is
$∴ Δ U = {nc}_{v} Δ T$
$Δ U = 5 \times \frac{R}{γ - 1} (T_{2} - T_{1}) (\begin{matrix} ∵ Specific heat, C_{v} = \frac{R}{γ - 1} \end{matrix})$
$= 5 \times \frac{8.3}{1.4 - 1} (673 - 273)$
$= 5 \times \frac{8.3}{0.4} \times 400$
$Δ U = 41.5 kJ$
So, the increase in internal energy of the gas is
$Δ U = 41.5 kJ$

AP EAMCET (20.04.2019) Shift-II

Thermodynamics

148441 In an adiabatic process, the pressure is increased by $\frac{2}{3} %$ .If $γ = \frac{2}{3}$ , then the volume decreases by nearly

1

\frac{4}{9} %

2

\frac{2}{3} %

3

1 %

4

\frac{9}{4} %

Explanation:

C For the adiabatic process.
${PV}^{γ} = Constant$
${PV}^{γ} = C$
Taking $\log$ on both sides.
Differentiating,
$\log P + γ \log V = \log C$
$\frac{Δ P}{P} + γ \frac{Δ V}{V} = 0$
$\frac{Δ V}{V} = \frac{- 1}{γ} (\frac{Δ P}{P})$
$\frac{Δ V}{V} \times 100 = \frac{- 1}{γ} (\frac{Δ P}{P} \times 100)$
$= - \frac{3}{2} \times \frac{2}{3}$
$= - 1 %$
So, volume decrease by $1 %$ .

BITSAT-2020

Thermodynamics

148442 Which equation is valid for adiabatic process?

1

{TV}^{γ - 1} =

constant

2

{pV}^{γ - 1} =

constant

3

T^{γ} V^{γ - 1} =

constant

4

\frac{p^{γ - 1}}{T^{γ - 1}} =

constant

Explanation:

A For an adiabatic process-
${PV}^{γ} =$ Constant
For an ideal gas $\frac{P V}{T} = n \bar{R} = C$
$P = \frac{C T}{V}$
$P = \frac{T}{V}$
From equation (i) and (ii)-
$(\frac{T}{V}) V^{γ} = constant$
$T . V^{γ - 1} = constant$

CG PET- 2014

Thermodynamics

148436 A hypothetical gas expands adiabatically such that its volume changes from 08 litres to 27 liters. If the ratio of final pressure of the gas to initial pressure of the gas is $\frac{16}{81}$ . Then the ratio of $\frac{c_{p}}{c_{v}}$ will be.

1

\frac{3}{1}

2

\frac{3}{2}

3

\frac{1}{2}

4

\frac{4}{3}

Explanation:

D We know, $\frac{c_{p}}{c_{v}} = γ$
For an adiabatic process,
${PV}^{γ} = C$
$P_{1} V_{1}^{γ} = P_{2} V_{2}^{γ}$
$\frac{P_{1}}{P_{2}} = {(\frac{V_{2}}{V_{1}})}^{γ}$
Given, $\frac{P_{2}}{P_{1}} = \frac{16}{81}$
$\frac{81}{16} = {(\frac{27}{8})}^{γ}$
${(\frac{3}{2})}^{4} = {(\frac{3}{2})}^{3 γ}$
By comparing,
$4 = 3 γ$
$\frac{c_{p}}{c_{v}} = γ = \frac{4}{3}$

JEE Main-31.01.2023

Thermodynamics

148439 The ratio of the adiabatic to isothermal elasticities of a triatomic (non-linear) gas is

1

3 : 4

2

1 : 2

3

4 : 3

4

5 : 3

Explanation:

C From elasticity of adiabatic process
$E_{adi} = γ P$
From elasticity of isothermal process
$E_{iso} = P$
Ratio of the adiabatic to isothermal elasticity,
$\frac{E_{adi}}{E_{iso}} = \frac{γ P}{P} = γ = 1 + \frac{2}{f} [For triatomic, f = 6]$
$= 1 + \frac{2}{6} = \frac{4}{3}$

AP EAMCET (21.09.2020) Shift-II

Thermodynamics

148440 Five moles of Hydrogen gas initially at STP is compressed adiabatically so that its temperature becomes $673 K$ . The increase in internal energy of the gas is $(R = 8.3 J {mol}^{- 1} K^{- 1}, γ = 1.4$ for diatomic gas)

1

80.5 kJ

2

21.55 kJ

3

41.50 kJ

4

65.55 kJ

Explanation:

C Final temperature $= 673 K = T_{2}$ Initial temperature $= 273 K = T_{1}$ Number of moles of hydrogen, $n = 5$ Universal gas constant $(R) = 8.3 J {mol}^{- 1} K^{- 1}$ For diatomic gas $(γ) = 1.4$
Now, change in internal energy due to adiabatic process is
$∴ Δ U = {nc}_{v} Δ T$
$Δ U = 5 \times \frac{R}{γ - 1} (T_{2} - T_{1}) (\begin{matrix} ∵ Specific heat, C_{v} = \frac{R}{γ - 1} \end{matrix})$
$= 5 \times \frac{8.3}{1.4 - 1} (673 - 273)$
$= 5 \times \frac{8.3}{0.4} \times 400$
$Δ U = 41.5 kJ$
So, the increase in internal energy of the gas is
$Δ U = 41.5 kJ$

AP EAMCET (20.04.2019) Shift-II

Thermodynamics

148441 In an adiabatic process, the pressure is increased by $\frac{2}{3} %$ .If $γ = \frac{2}{3}$ , then the volume decreases by nearly

1

\frac{4}{9} %

2

\frac{2}{3} %

3

1 %

4

\frac{9}{4} %

Explanation:

C For the adiabatic process.
${PV}^{γ} = Constant$
${PV}^{γ} = C$
Taking $\log$ on both sides.
Differentiating,
$\log P + γ \log V = \log C$
$\frac{Δ P}{P} + γ \frac{Δ V}{V} = 0$
$\frac{Δ V}{V} = \frac{- 1}{γ} (\frac{Δ P}{P})$
$\frac{Δ V}{V} \times 100 = \frac{- 1}{γ} (\frac{Δ P}{P} \times 100)$
$= - \frac{3}{2} \times \frac{2}{3}$
$= - 1 %$
So, volume decrease by $1 %$ .

BITSAT-2020

Thermodynamics

148442 Which equation is valid for adiabatic process?

1

{TV}^{γ - 1} =

constant

2

{pV}^{γ - 1} =

constant

3

T^{γ} V^{γ - 1} =

constant

4

\frac{p^{γ - 1}}{T^{γ - 1}} =

constant

Explanation:

A For an adiabatic process-
${PV}^{γ} =$ Constant
For an ideal gas $\frac{P V}{T} = n \bar{R} = C$
$P = \frac{C T}{V}$
$P = \frac{T}{V}$
From equation (i) and (ii)-
$(\frac{T}{V}) V^{γ} = constant$
$T . V^{γ - 1} = constant$

CG PET- 2014

Thermodynamics

148436 A hypothetical gas expands adiabatically such that its volume changes from 08 litres to 27 liters. If the ratio of final pressure of the gas to initial pressure of the gas is $\frac{16}{81}$ . Then the ratio of $\frac{c_{p}}{c_{v}}$ will be.

1

\frac{3}{1}

2

\frac{3}{2}

3

\frac{1}{2}

4

\frac{4}{3}

Explanation:

D We know, $\frac{c_{p}}{c_{v}} = γ$
For an adiabatic process,
${PV}^{γ} = C$
$P_{1} V_{1}^{γ} = P_{2} V_{2}^{γ}$
$\frac{P_{1}}{P_{2}} = {(\frac{V_{2}}{V_{1}})}^{γ}$
Given, $\frac{P_{2}}{P_{1}} = \frac{16}{81}$
$\frac{81}{16} = {(\frac{27}{8})}^{γ}$
${(\frac{3}{2})}^{4} = {(\frac{3}{2})}^{3 γ}$
By comparing,
$4 = 3 γ$
$\frac{c_{p}}{c_{v}} = γ = \frac{4}{3}$

JEE Main-31.01.2023

Thermodynamics

148439 The ratio of the adiabatic to isothermal elasticities of a triatomic (non-linear) gas is

1

3 : 4

2

1 : 2

3

4 : 3

4

5 : 3

Explanation:

C From elasticity of adiabatic process
$E_{adi} = γ P$
From elasticity of isothermal process
$E_{iso} = P$
Ratio of the adiabatic to isothermal elasticity,
$\frac{E_{adi}}{E_{iso}} = \frac{γ P}{P} = γ = 1 + \frac{2}{f} [For triatomic, f = 6]$
$= 1 + \frac{2}{6} = \frac{4}{3}$

AP EAMCET (21.09.2020) Shift-II

Thermodynamics

148440 Five moles of Hydrogen gas initially at STP is compressed adiabatically so that its temperature becomes $673 K$ . The increase in internal energy of the gas is $(R = 8.3 J {mol}^{- 1} K^{- 1}, γ = 1.4$ for diatomic gas)

1

80.5 kJ

2

21.55 kJ

3

41.50 kJ

4

65.55 kJ

Explanation:

C Final temperature $= 673 K = T_{2}$ Initial temperature $= 273 K = T_{1}$ Number of moles of hydrogen, $n = 5$ Universal gas constant $(R) = 8.3 J {mol}^{- 1} K^{- 1}$ For diatomic gas $(γ) = 1.4$
Now, change in internal energy due to adiabatic process is
$∴ Δ U = {nc}_{v} Δ T$
$Δ U = 5 \times \frac{R}{γ - 1} (T_{2} - T_{1}) (\begin{matrix} ∵ Specific heat, C_{v} = \frac{R}{γ - 1} \end{matrix})$
$= 5 \times \frac{8.3}{1.4 - 1} (673 - 273)$
$= 5 \times \frac{8.3}{0.4} \times 400$
$Δ U = 41.5 kJ$
So, the increase in internal energy of the gas is
$Δ U = 41.5 kJ$

AP EAMCET (20.04.2019) Shift-II

Thermodynamics

148441 In an adiabatic process, the pressure is increased by $\frac{2}{3} %$ .If $γ = \frac{2}{3}$ , then the volume decreases by nearly

1

\frac{4}{9} %

2

\frac{2}{3} %

3

1 %

4

\frac{9}{4} %

Explanation:

C For the adiabatic process.
${PV}^{γ} = Constant$
${PV}^{γ} = C$
Taking $\log$ on both sides.
Differentiating,
$\log P + γ \log V = \log C$
$\frac{Δ P}{P} + γ \frac{Δ V}{V} = 0$
$\frac{Δ V}{V} = \frac{- 1}{γ} (\frac{Δ P}{P})$
$\frac{Δ V}{V} \times 100 = \frac{- 1}{γ} (\frac{Δ P}{P} \times 100)$
$= - \frac{3}{2} \times \frac{2}{3}$
$= - 1 %$
So, volume decrease by $1 %$ .

BITSAT-2020

Thermodynamics

148442 Which equation is valid for adiabatic process?

1

{TV}^{γ - 1} =

constant

2

{pV}^{γ - 1} =

constant

3

T^{γ} V^{γ - 1} =

constant

4

\frac{p^{γ - 1}}{T^{γ - 1}} =

constant

Explanation:

A For an adiabatic process-
${PV}^{γ} =$ Constant
For an ideal gas $\frac{P V}{T} = n \bar{R} = C$
$P = \frac{C T}{V}$
$P = \frac{T}{V}$
From equation (i) and (ii)-
$(\frac{T}{V}) V^{γ} = constant$
$T . V^{γ - 1} = constant$

CG PET- 2014

Thermodynamics

148436 A hypothetical gas expands adiabatically such that its volume changes from 08 litres to 27 liters. If the ratio of final pressure of the gas to initial pressure of the gas is $\frac{16}{81}$ . Then the ratio of $\frac{c_{p}}{c_{v}}$ will be.

1

\frac{3}{1}

2

\frac{3}{2}

3

\frac{1}{2}

4

\frac{4}{3}

Explanation:

D We know, $\frac{c_{p}}{c_{v}} = γ$
For an adiabatic process,
${PV}^{γ} = C$
$P_{1} V_{1}^{γ} = P_{2} V_{2}^{γ}$
$\frac{P_{1}}{P_{2}} = {(\frac{V_{2}}{V_{1}})}^{γ}$
Given, $\frac{P_{2}}{P_{1}} = \frac{16}{81}$
$\frac{81}{16} = {(\frac{27}{8})}^{γ}$
${(\frac{3}{2})}^{4} = {(\frac{3}{2})}^{3 γ}$
By comparing,
$4 = 3 γ$
$\frac{c_{p}}{c_{v}} = γ = \frac{4}{3}$

JEE Main-31.01.2023

Thermodynamics

148439 The ratio of the adiabatic to isothermal elasticities of a triatomic (non-linear) gas is

1

3 : 4

2

1 : 2

3

4 : 3

4

5 : 3

Explanation:

C From elasticity of adiabatic process
$E_{adi} = γ P$
From elasticity of isothermal process
$E_{iso} = P$
Ratio of the adiabatic to isothermal elasticity,
$\frac{E_{adi}}{E_{iso}} = \frac{γ P}{P} = γ = 1 + \frac{2}{f} [For triatomic, f = 6]$
$= 1 + \frac{2}{6} = \frac{4}{3}$

AP EAMCET (21.09.2020) Shift-II

Thermodynamics

148440 Five moles of Hydrogen gas initially at STP is compressed adiabatically so that its temperature becomes $673 K$ . The increase in internal energy of the gas is $(R = 8.3 J {mol}^{- 1} K^{- 1}, γ = 1.4$ for diatomic gas)

1

80.5 kJ

2

21.55 kJ

3

41.50 kJ

4

65.55 kJ

Explanation:

C Final temperature $= 673 K = T_{2}$ Initial temperature $= 273 K = T_{1}$ Number of moles of hydrogen, $n = 5$ Universal gas constant $(R) = 8.3 J {mol}^{- 1} K^{- 1}$ For diatomic gas $(γ) = 1.4$
Now, change in internal energy due to adiabatic process is
$∴ Δ U = {nc}_{v} Δ T$
$Δ U = 5 \times \frac{R}{γ - 1} (T_{2} - T_{1}) (\begin{matrix} ∵ Specific heat, C_{v} = \frac{R}{γ - 1} \end{matrix})$
$= 5 \times \frac{8.3}{1.4 - 1} (673 - 273)$
$= 5 \times \frac{8.3}{0.4} \times 400$
$Δ U = 41.5 kJ$
So, the increase in internal energy of the gas is
$Δ U = 41.5 kJ$

AP EAMCET (20.04.2019) Shift-II

Thermodynamics

148441 In an adiabatic process, the pressure is increased by $\frac{2}{3} %$ .If $γ = \frac{2}{3}$ , then the volume decreases by nearly

1

\frac{4}{9} %

2

\frac{2}{3} %

3

1 %

4

\frac{9}{4} %

Explanation:

C For the adiabatic process.
${PV}^{γ} = Constant$
${PV}^{γ} = C$
Taking $\log$ on both sides.
Differentiating,
$\log P + γ \log V = \log C$
$\frac{Δ P}{P} + γ \frac{Δ V}{V} = 0$
$\frac{Δ V}{V} = \frac{- 1}{γ} (\frac{Δ P}{P})$
$\frac{Δ V}{V} \times 100 = \frac{- 1}{γ} (\frac{Δ P}{P} \times 100)$
$= - \frac{3}{2} \times \frac{2}{3}$
$= - 1 %$
So, volume decrease by $1 %$ .

BITSAT-2020

Thermodynamics

148442 Which equation is valid for adiabatic process?

1

{TV}^{γ - 1} =

constant

2

{pV}^{γ - 1} =

constant

3

T^{γ} V^{γ - 1} =

constant

4

\frac{p^{γ - 1}}{T^{γ - 1}} =

constant

Explanation:

A For an adiabatic process-
${PV}^{γ} =$ Constant
For an ideal gas $\frac{P V}{T} = n \bar{R} = C$
$P = \frac{C T}{V}$
$P = \frac{T}{V}$
From equation (i) and (ii)-
$(\frac{T}{V}) V^{γ} = constant$
$T . V^{γ - 1} = constant$

CG PET- 2014

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