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Thermodynamics

148256 A gas is suddenly expanded such that its final volume becomes 3 times its initial volume. If the specific heat at constant volume of the gas is $2 R$ , then the ratio of initial to final pressures is nearly equal to

1 5

2 6.5

3 7

4 3.5

Explanation:

A Given,
$V_{1} = V, V_{2} = 3 V, C_{V} = 2 R, C_{P} - C_{V} = R$ ,
$C_{P} = R + C_{V} = R + 2 R = 3 R$
$γ = \frac{C_{P}}{C_{V}} = \frac{3 R}{2 R} = 1.5 C_{P} = 3 R$
A gas is suddenly expanded, so adiabatic process
${PV}^{γ} =$ const
$P_{1} V_{1}^{γ} = P_{2} V_{2}^{γ}$
$(\frac{P_{1}}{P_{2}}) = {(\frac{V_{2}}{V_{1}})}^{1.5} = {(\frac{3 V}{V})}^{1.5}$
$P_{1} = P_{2} \times (3)^{1.5} = P_{2} \times 5.196$
$P_{1} = 5.19 P_{2}$
$\frac{P_{1}}{P_{2}} = 5.19 \approx 5$

Manipal UGET-2010

Thermodynamics

148257 An ideal gas expands along the path $AB$ as shown in the p-V diagram. The work done is

1

4 J

2

1.2 J

3

2.4 J

4 None of these

Explanation:

B
Net workdone $=$ Area under $P - V$ graph
$=$ Area of $ABCD$
$= \frac{1}{2} (AD + BC) \times CD$
$= \frac{1}{2} (4 + 8) \times 0.2$
$= 12 \times 0.1 = 1.2$ Joule

Manipal UGET-2010

Thermodynamics

148259 One mole of nitrogen gas being initially at a temperature of $T_{0} = 300 K$ is adiabatically compressed to increase its pressure 10 times. The final gas temperature after compression is (Assume, nitrogen gas molecules as rigid diatomic and $100^{1 / 7} = 1.9$ )

1

120 K

2

750 K

3

650 K

4

570 K

Explanation:

D We know that,
$N_{2}$ gas is the diatomic gas,
$γ = \frac{7}{5} = 1.4$
$T_{0} = 300 K, P_{1} = P_{0}, P_{2} = 10 \times P_{0}$
For adiabatic process,
$P^{1 - γ} T^{γ} =$ const
$P_{1}^{1 - γ} T_{1}^{γ} = P_{2}^{1 - γ} T_{2}^{γ}$
${(\frac{T_{2}}{T_{1}})}^{γ} = {(\frac{P_{2}}{P_{1}})}^{γ - 1}$
$\frac{T_{2}}{T_{1}} = {(\frac{P_{2}}{P_{1}})}^{\frac{γ - 1}{γ}} = {(\frac{10 . P_{0}}{P_{o}})}^{\frac{1.4 - 1}{1.4}}$
$\frac{T_{2}}{T_{1}} = (10)^{\frac{0.4}{1.4}} = (100)^{1 / 7} = 1.9$
$T_{2} = 1.9 \times T_{1} = 1.9 \times 300 = 570 K$
$T_{2} = 570 K$
$∴$ Final gas temperature after compression $= 570 K$

TS- EAMCET-03.05.2019

Thermodynamics

148260 A diesel engine has a compression ratio of 20 :1. If the initial pressure is $1 \times 10^{5} Pa$ and the initial volume of the cylinder is $1 \times 10^{- 3} m^{3}$ , then how much work does the gas do during the compression?
(Assume the process as adiabatic)
$(C_{v} = 20.8 J / mol K, γ_{air} = 1.4, (20)^{1.4} = 66.3)$

1

- 880 J

2

- 579 J

3

220 J

4

485 J

Explanation:

B Given,
Compression ratio $= \frac{Initial volume (V_{i})}{Final volume (V_{f})} = \frac{20}{1}$
$V_{i} = 20 \times V_{f}$
...(i)
$P_{i} = 1 \times 10^{5} Pa, V_{i} = 1 \times 10^{- 3} m^{3}$
Putting $V_{i}$ value is equation (i) we get,
$1 \times 10^{- 3} = 20 V_{f} \Rightarrow V_{f} = \frac{1}{20} \times 10^{- 3} m^{3}$
$= 0.05 \times 10^{- 3} m^{3}$
For adiabatic process,
${PV}^{γ} =$ const, $P_{i} V_{i}^{γ} = P_{f} V_{f}^{γ}$
$P_{f} = P_{i} {(\frac{V_{i}}{V_{f}})}^{γ}$
$= P_{i} \times (20)^{1.4} = 10^{5} \times 66.289$
$P_{f} = 66.3 \times 10^{5} Pa$
$W = \frac{P_{i} V_{i} - P_{f} V_{f}}{γ - 1} = \frac{(10^{5} \times 10^{- 3}) - (66.3 \times 10^{5} \times 0.05 \times 10^{- 3})}{1.4 - 1}$
$= \frac{10^{2} - 3.31 \times 10^{2}}{0.4}$
$W = \frac{- 2.315 \times 10^{2}}{0.4} = - 5.7875 \times 10^{2}$
$W = - 578.7 J \approx - 579 J$

TS- EAMCET-07.05.2018

Thermodynamics

148256 A gas is suddenly expanded such that its final volume becomes 3 times its initial volume. If the specific heat at constant volume of the gas is $2 R$ , then the ratio of initial to final pressures is nearly equal to

1 5

2 6.5

3 7

4 3.5

Explanation:

A Given,
$V_{1} = V, V_{2} = 3 V, C_{V} = 2 R, C_{P} - C_{V} = R$ ,
$C_{P} = R + C_{V} = R + 2 R = 3 R$
$γ = \frac{C_{P}}{C_{V}} = \frac{3 R}{2 R} = 1.5 C_{P} = 3 R$
A gas is suddenly expanded, so adiabatic process
${PV}^{γ} =$ const
$P_{1} V_{1}^{γ} = P_{2} V_{2}^{γ}$
$(\frac{P_{1}}{P_{2}}) = {(\frac{V_{2}}{V_{1}})}^{1.5} = {(\frac{3 V}{V})}^{1.5}$
$P_{1} = P_{2} \times (3)^{1.5} = P_{2} \times 5.196$
$P_{1} = 5.19 P_{2}$
$\frac{P_{1}}{P_{2}} = 5.19 \approx 5$

Manipal UGET-2010

Thermodynamics

148257 An ideal gas expands along the path $AB$ as shown in the p-V diagram. The work done is

1

4 J

2

1.2 J

3

2.4 J

4 None of these

Explanation:

B
Net workdone $=$ Area under $P - V$ graph
$=$ Area of $ABCD$
$= \frac{1}{2} (AD + BC) \times CD$
$= \frac{1}{2} (4 + 8) \times 0.2$
$= 12 \times 0.1 = 1.2$ Joule

Manipal UGET-2010

Thermodynamics

148259 One mole of nitrogen gas being initially at a temperature of $T_{0} = 300 K$ is adiabatically compressed to increase its pressure 10 times. The final gas temperature after compression is (Assume, nitrogen gas molecules as rigid diatomic and $100^{1 / 7} = 1.9$ )

1

120 K

2

750 K

3

650 K

4

570 K

Explanation:

D We know that,
$N_{2}$ gas is the diatomic gas,
$γ = \frac{7}{5} = 1.4$
$T_{0} = 300 K, P_{1} = P_{0}, P_{2} = 10 \times P_{0}$
For adiabatic process,
$P^{1 - γ} T^{γ} =$ const
$P_{1}^{1 - γ} T_{1}^{γ} = P_{2}^{1 - γ} T_{2}^{γ}$
${(\frac{T_{2}}{T_{1}})}^{γ} = {(\frac{P_{2}}{P_{1}})}^{γ - 1}$
$\frac{T_{2}}{T_{1}} = {(\frac{P_{2}}{P_{1}})}^{\frac{γ - 1}{γ}} = {(\frac{10 . P_{0}}{P_{o}})}^{\frac{1.4 - 1}{1.4}}$
$\frac{T_{2}}{T_{1}} = (10)^{\frac{0.4}{1.4}} = (100)^{1 / 7} = 1.9$
$T_{2} = 1.9 \times T_{1} = 1.9 \times 300 = 570 K$
$T_{2} = 570 K$
$∴$ Final gas temperature after compression $= 570 K$

TS- EAMCET-03.05.2019

Thermodynamics

148260 A diesel engine has a compression ratio of 20 :1. If the initial pressure is $1 \times 10^{5} Pa$ and the initial volume of the cylinder is $1 \times 10^{- 3} m^{3}$ , then how much work does the gas do during the compression?
(Assume the process as adiabatic)
$(C_{v} = 20.8 J / mol K, γ_{air} = 1.4, (20)^{1.4} = 66.3)$

1

- 880 J

2

- 579 J

3

220 J

4

485 J

Explanation:

B Given,
Compression ratio $= \frac{Initial volume (V_{i})}{Final volume (V_{f})} = \frac{20}{1}$
$V_{i} = 20 \times V_{f}$
...(i)
$P_{i} = 1 \times 10^{5} Pa, V_{i} = 1 \times 10^{- 3} m^{3}$
Putting $V_{i}$ value is equation (i) we get,
$1 \times 10^{- 3} = 20 V_{f} \Rightarrow V_{f} = \frac{1}{20} \times 10^{- 3} m^{3}$
$= 0.05 \times 10^{- 3} m^{3}$
For adiabatic process,
${PV}^{γ} =$ const, $P_{i} V_{i}^{γ} = P_{f} V_{f}^{γ}$
$P_{f} = P_{i} {(\frac{V_{i}}{V_{f}})}^{γ}$
$= P_{i} \times (20)^{1.4} = 10^{5} \times 66.289$
$P_{f} = 66.3 \times 10^{5} Pa$
$W = \frac{P_{i} V_{i} - P_{f} V_{f}}{γ - 1} = \frac{(10^{5} \times 10^{- 3}) - (66.3 \times 10^{5} \times 0.05 \times 10^{- 3})}{1.4 - 1}$
$= \frac{10^{2} - 3.31 \times 10^{2}}{0.4}$
$W = \frac{- 2.315 \times 10^{2}}{0.4} = - 5.7875 \times 10^{2}$
$W = - 578.7 J \approx - 579 J$

TS- EAMCET-07.05.2018

Thermodynamics

148256 A gas is suddenly expanded such that its final volume becomes 3 times its initial volume. If the specific heat at constant volume of the gas is $2 R$ , then the ratio of initial to final pressures is nearly equal to

1 5

2 6.5

3 7

4 3.5

Explanation:

A Given,
$V_{1} = V, V_{2} = 3 V, C_{V} = 2 R, C_{P} - C_{V} = R$ ,
$C_{P} = R + C_{V} = R + 2 R = 3 R$
$γ = \frac{C_{P}}{C_{V}} = \frac{3 R}{2 R} = 1.5 C_{P} = 3 R$
A gas is suddenly expanded, so adiabatic process
${PV}^{γ} =$ const
$P_{1} V_{1}^{γ} = P_{2} V_{2}^{γ}$
$(\frac{P_{1}}{P_{2}}) = {(\frac{V_{2}}{V_{1}})}^{1.5} = {(\frac{3 V}{V})}^{1.5}$
$P_{1} = P_{2} \times (3)^{1.5} = P_{2} \times 5.196$
$P_{1} = 5.19 P_{2}$
$\frac{P_{1}}{P_{2}} = 5.19 \approx 5$

Manipal UGET-2010

Thermodynamics

148257 An ideal gas expands along the path $AB$ as shown in the p-V diagram. The work done is

1

4 J

2

1.2 J

3

2.4 J

4 None of these

Explanation:

B
Net workdone $=$ Area under $P - V$ graph
$=$ Area of $ABCD$
$= \frac{1}{2} (AD + BC) \times CD$
$= \frac{1}{2} (4 + 8) \times 0.2$
$= 12 \times 0.1 = 1.2$ Joule

Manipal UGET-2010

Thermodynamics

148259 One mole of nitrogen gas being initially at a temperature of $T_{0} = 300 K$ is adiabatically compressed to increase its pressure 10 times. The final gas temperature after compression is (Assume, nitrogen gas molecules as rigid diatomic and $100^{1 / 7} = 1.9$ )

1

120 K

2

750 K

3

650 K

4

570 K

Explanation:

D We know that,
$N_{2}$ gas is the diatomic gas,
$γ = \frac{7}{5} = 1.4$
$T_{0} = 300 K, P_{1} = P_{0}, P_{2} = 10 \times P_{0}$
For adiabatic process,
$P^{1 - γ} T^{γ} =$ const
$P_{1}^{1 - γ} T_{1}^{γ} = P_{2}^{1 - γ} T_{2}^{γ}$
${(\frac{T_{2}}{T_{1}})}^{γ} = {(\frac{P_{2}}{P_{1}})}^{γ - 1}$
$\frac{T_{2}}{T_{1}} = {(\frac{P_{2}}{P_{1}})}^{\frac{γ - 1}{γ}} = {(\frac{10 . P_{0}}{P_{o}})}^{\frac{1.4 - 1}{1.4}}$
$\frac{T_{2}}{T_{1}} = (10)^{\frac{0.4}{1.4}} = (100)^{1 / 7} = 1.9$
$T_{2} = 1.9 \times T_{1} = 1.9 \times 300 = 570 K$
$T_{2} = 570 K$
$∴$ Final gas temperature after compression $= 570 K$

TS- EAMCET-03.05.2019

Thermodynamics

148260 A diesel engine has a compression ratio of 20 :1. If the initial pressure is $1 \times 10^{5} Pa$ and the initial volume of the cylinder is $1 \times 10^{- 3} m^{3}$ , then how much work does the gas do during the compression?
(Assume the process as adiabatic)
$(C_{v} = 20.8 J / mol K, γ_{air} = 1.4, (20)^{1.4} = 66.3)$

1

- 880 J

2

- 579 J

3

220 J

4

485 J

Explanation:

B Given,
Compression ratio $= \frac{Initial volume (V_{i})}{Final volume (V_{f})} = \frac{20}{1}$
$V_{i} = 20 \times V_{f}$
...(i)
$P_{i} = 1 \times 10^{5} Pa, V_{i} = 1 \times 10^{- 3} m^{3}$
Putting $V_{i}$ value is equation (i) we get,
$1 \times 10^{- 3} = 20 V_{f} \Rightarrow V_{f} = \frac{1}{20} \times 10^{- 3} m^{3}$
$= 0.05 \times 10^{- 3} m^{3}$
For adiabatic process,
${PV}^{γ} =$ const, $P_{i} V_{i}^{γ} = P_{f} V_{f}^{γ}$
$P_{f} = P_{i} {(\frac{V_{i}}{V_{f}})}^{γ}$
$= P_{i} \times (20)^{1.4} = 10^{5} \times 66.289$
$P_{f} = 66.3 \times 10^{5} Pa$
$W = \frac{P_{i} V_{i} - P_{f} V_{f}}{γ - 1} = \frac{(10^{5} \times 10^{- 3}) - (66.3 \times 10^{5} \times 0.05 \times 10^{- 3})}{1.4 - 1}$
$= \frac{10^{2} - 3.31 \times 10^{2}}{0.4}$
$W = \frac{- 2.315 \times 10^{2}}{0.4} = - 5.7875 \times 10^{2}$
$W = - 578.7 J \approx - 579 J$

TS- EAMCET-07.05.2018

Thermodynamics

148256 A gas is suddenly expanded such that its final volume becomes 3 times its initial volume. If the specific heat at constant volume of the gas is $2 R$ , then the ratio of initial to final pressures is nearly equal to

1 5

2 6.5

3 7

4 3.5

Explanation:

A Given,
$V_{1} = V, V_{2} = 3 V, C_{V} = 2 R, C_{P} - C_{V} = R$ ,
$C_{P} = R + C_{V} = R + 2 R = 3 R$
$γ = \frac{C_{P}}{C_{V}} = \frac{3 R}{2 R} = 1.5 C_{P} = 3 R$
A gas is suddenly expanded, so adiabatic process
${PV}^{γ} =$ const
$P_{1} V_{1}^{γ} = P_{2} V_{2}^{γ}$
$(\frac{P_{1}}{P_{2}}) = {(\frac{V_{2}}{V_{1}})}^{1.5} = {(\frac{3 V}{V})}^{1.5}$
$P_{1} = P_{2} \times (3)^{1.5} = P_{2} \times 5.196$
$P_{1} = 5.19 P_{2}$
$\frac{P_{1}}{P_{2}} = 5.19 \approx 5$

Manipal UGET-2010

Thermodynamics

148257 An ideal gas expands along the path $AB$ as shown in the p-V diagram. The work done is

1

4 J

2

1.2 J

3

2.4 J

4 None of these

Explanation:

B
Net workdone $=$ Area under $P - V$ graph
$=$ Area of $ABCD$
$= \frac{1}{2} (AD + BC) \times CD$
$= \frac{1}{2} (4 + 8) \times 0.2$
$= 12 \times 0.1 = 1.2$ Joule

Manipal UGET-2010

Thermodynamics

148259 One mole of nitrogen gas being initially at a temperature of $T_{0} = 300 K$ is adiabatically compressed to increase its pressure 10 times. The final gas temperature after compression is (Assume, nitrogen gas molecules as rigid diatomic and $100^{1 / 7} = 1.9$ )

1

120 K

2

750 K

3

650 K

4

570 K

Explanation:

D We know that,
$N_{2}$ gas is the diatomic gas,
$γ = \frac{7}{5} = 1.4$
$T_{0} = 300 K, P_{1} = P_{0}, P_{2} = 10 \times P_{0}$
For adiabatic process,
$P^{1 - γ} T^{γ} =$ const
$P_{1}^{1 - γ} T_{1}^{γ} = P_{2}^{1 - γ} T_{2}^{γ}$
${(\frac{T_{2}}{T_{1}})}^{γ} = {(\frac{P_{2}}{P_{1}})}^{γ - 1}$
$\frac{T_{2}}{T_{1}} = {(\frac{P_{2}}{P_{1}})}^{\frac{γ - 1}{γ}} = {(\frac{10 . P_{0}}{P_{o}})}^{\frac{1.4 - 1}{1.4}}$
$\frac{T_{2}}{T_{1}} = (10)^{\frac{0.4}{1.4}} = (100)^{1 / 7} = 1.9$
$T_{2} = 1.9 \times T_{1} = 1.9 \times 300 = 570 K$
$T_{2} = 570 K$
$∴$ Final gas temperature after compression $= 570 K$

TS- EAMCET-03.05.2019

Thermodynamics

148260 A diesel engine has a compression ratio of 20 :1. If the initial pressure is $1 \times 10^{5} Pa$ and the initial volume of the cylinder is $1 \times 10^{- 3} m^{3}$ , then how much work does the gas do during the compression?
(Assume the process as adiabatic)
$(C_{v} = 20.8 J / mol K, γ_{air} = 1.4, (20)^{1.4} = 66.3)$

1

- 880 J

2

- 579 J

3

220 J

4

485 J

Explanation:

B Given,
Compression ratio $= \frac{Initial volume (V_{i})}{Final volume (V_{f})} = \frac{20}{1}$
$V_{i} = 20 \times V_{f}$
...(i)
$P_{i} = 1 \times 10^{5} Pa, V_{i} = 1 \times 10^{- 3} m^{3}$
Putting $V_{i}$ value is equation (i) we get,
$1 \times 10^{- 3} = 20 V_{f} \Rightarrow V_{f} = \frac{1}{20} \times 10^{- 3} m^{3}$
$= 0.05 \times 10^{- 3} m^{3}$
For adiabatic process,
${PV}^{γ} =$ const, $P_{i} V_{i}^{γ} = P_{f} V_{f}^{γ}$
$P_{f} = P_{i} {(\frac{V_{i}}{V_{f}})}^{γ}$
$= P_{i} \times (20)^{1.4} = 10^{5} \times 66.289$
$P_{f} = 66.3 \times 10^{5} Pa$
$W = \frac{P_{i} V_{i} - P_{f} V_{f}}{γ - 1} = \frac{(10^{5} \times 10^{- 3}) - (66.3 \times 10^{5} \times 0.05 \times 10^{- 3})}{1.4 - 1}$
$= \frac{10^{2} - 3.31 \times 10^{2}}{0.4}$
$W = \frac{- 2.315 \times 10^{2}}{0.4} = - 5.7875 \times 10^{2}$
$W = - 578.7 J \approx - 579 J$

TS- EAMCET-07.05.2018

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