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Thermodynamics

148378 A diatomic gas $(C_{p} = \frac{7}{2} R)$ does $200 J$ of work when it is expanded isobarically. The heat given to the gas in the process is

1

600 J

2

800 J

3

900 J

4

700 J

Explanation:

D Given,
$\frac{C_{P}}{C_{V}} = γ \Rightarrow C_{V} = \frac{C_{P}}{γ}$
$C_{P} = \frac{7}{2} R, For diatomic$
$C_{v} = \frac{7 R}{2 γ} = \frac{7 R}{2 \times 1.4}$
We know that,
From $1^{st}$ law of thermodynamics
$dQ = dU + dW$
$dQ = {mC}_{v} Δ T + dW$
$= m \times \frac{7 R}{2 \times 1.4} Δ T + PdV$
$= (\frac{70}{2 \times 14}) PdV + PdV$
$= \frac{7}{2} PdV$
$dQ = \frac{7}{2} \times 200$
$dQ = 700 J$

TS EAMCET 18.07.2022

Thermodynamics

148379 A monoatomic gas does $100 J$ of work when it is expanded isobarically. How much of heat is given to the gas in the process

1

150 J

2

200 J

3

250 J

4

300 J

Explanation:

C Given,
Gas is mono atomic
$∴ ∴ γ = \frac{5}{3}$
$C_{P} - C_{V} = R [Mayer's equation]$
$\frac{C_{P}}{C_{V}} = γ$
$C_{P} = γ C_{V}$
$C_{V} (γ - 1) = R$
$C_{V} = \frac{R}{γ - 1}$
Now from $1^{st}$ Law of thermodynamics
$dQ = dU + dW$
$= {mC}_{V} dT + PdV$
$= \frac{m \times RdT}{γ - 1} + PdV$
$dQ = \frac{PdV}{\frac{5}{3} - 1} + PdV$
$= \frac{3}{2} PdV + PdV$
$dQ = \frac{5}{2} \times PdV$
$dQ = \frac{5}{2} \times 100$
$dQ = 250 J$

TS EAMCET 18.07.2022

Thermodynamics

148380 If the temperature of $50 ml$ of gas at $27^{\circ} C$ raised to $57^{\circ} C$ at constant pressure, the final volume of the gas is

1

23.7 ml

2

25.0 ml

3

53.7 ml

4

55.0 ml

Explanation:

D Given,
Initial volume $(V_{1}) = 50 ml$
Initial temperature $(T_{1}) = 27^{\circ} C = 300 K$
Final Temperature $(T_{2}) = 57^{\circ} C = 330 K$
Ideal gas equation
$PV = nRT$
For constant pressure process
$\frac{V}{T} =$ Constant
$∴ \frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$
$\frac{50}{300} = \frac{V_{2}}{330}$
$V_{2} = 55 ml$

TS EAMCET 03.05.2018

Thermodynamics

148381 A thermodynamic system is taken from an original state $D$ to an intermediate state $E$ by linear process shown in the figure. Its volume is then reduced to the original volume from $E$ to $F$ by an isobaric process. The total work done by the gas from $D$ to $E$ to $F$ will be.

1

- 450 J

2

450 J

3

900 J

4

1350 J

Explanation:

B Given

We know that,
Area under the $P - V$ diagram is equal to work done.
$Δ W = \frac{1}{2} \times 300 \times 3 = 450 J$

JEE Main-29.07.2022

Thermodynamics

148378 A diatomic gas $(C_{p} = \frac{7}{2} R)$ does $200 J$ of work when it is expanded isobarically. The heat given to the gas in the process is

1

600 J

2

800 J

3

900 J

4

700 J

Explanation:

D Given,
$\frac{C_{P}}{C_{V}} = γ \Rightarrow C_{V} = \frac{C_{P}}{γ}$
$C_{P} = \frac{7}{2} R, For diatomic$
$C_{v} = \frac{7 R}{2 γ} = \frac{7 R}{2 \times 1.4}$
We know that,
From $1^{st}$ law of thermodynamics
$dQ = dU + dW$
$dQ = {mC}_{v} Δ T + dW$
$= m \times \frac{7 R}{2 \times 1.4} Δ T + PdV$
$= (\frac{70}{2 \times 14}) PdV + PdV$
$= \frac{7}{2} PdV$
$dQ = \frac{7}{2} \times 200$
$dQ = 700 J$

TS EAMCET 18.07.2022

Thermodynamics

148379 A monoatomic gas does $100 J$ of work when it is expanded isobarically. How much of heat is given to the gas in the process

1

150 J

2

200 J

3

250 J

4

300 J

Explanation:

C Given,
Gas is mono atomic
$∴ ∴ γ = \frac{5}{3}$
$C_{P} - C_{V} = R [Mayer's equation]$
$\frac{C_{P}}{C_{V}} = γ$
$C_{P} = γ C_{V}$
$C_{V} (γ - 1) = R$
$C_{V} = \frac{R}{γ - 1}$
Now from $1^{st}$ Law of thermodynamics
$dQ = dU + dW$
$= {mC}_{V} dT + PdV$
$= \frac{m \times RdT}{γ - 1} + PdV$
$dQ = \frac{PdV}{\frac{5}{3} - 1} + PdV$
$= \frac{3}{2} PdV + PdV$
$dQ = \frac{5}{2} \times PdV$
$dQ = \frac{5}{2} \times 100$
$dQ = 250 J$

TS EAMCET 18.07.2022

Thermodynamics

148380 If the temperature of $50 ml$ of gas at $27^{\circ} C$ raised to $57^{\circ} C$ at constant pressure, the final volume of the gas is

1

23.7 ml

2

25.0 ml

3

53.7 ml

4

55.0 ml

Explanation:

D Given,
Initial volume $(V_{1}) = 50 ml$
Initial temperature $(T_{1}) = 27^{\circ} C = 300 K$
Final Temperature $(T_{2}) = 57^{\circ} C = 330 K$
Ideal gas equation
$PV = nRT$
For constant pressure process
$\frac{V}{T} =$ Constant
$∴ \frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$
$\frac{50}{300} = \frac{V_{2}}{330}$
$V_{2} = 55 ml$

TS EAMCET 03.05.2018

Thermodynamics

148381 A thermodynamic system is taken from an original state $D$ to an intermediate state $E$ by linear process shown in the figure. Its volume is then reduced to the original volume from $E$ to $F$ by an isobaric process. The total work done by the gas from $D$ to $E$ to $F$ will be.

1

- 450 J

2

450 J

3

900 J

4

1350 J

Explanation:

B Given

We know that,
Area under the $P - V$ diagram is equal to work done.
$Δ W = \frac{1}{2} \times 300 \times 3 = 450 J$

JEE Main-29.07.2022

Thermodynamics

148378 A diatomic gas $(C_{p} = \frac{7}{2} R)$ does $200 J$ of work when it is expanded isobarically. The heat given to the gas in the process is

1

600 J

2

800 J

3

900 J

4

700 J

Explanation:

D Given,
$\frac{C_{P}}{C_{V}} = γ \Rightarrow C_{V} = \frac{C_{P}}{γ}$
$C_{P} = \frac{7}{2} R, For diatomic$
$C_{v} = \frac{7 R}{2 γ} = \frac{7 R}{2 \times 1.4}$
We know that,
From $1^{st}$ law of thermodynamics
$dQ = dU + dW$
$dQ = {mC}_{v} Δ T + dW$
$= m \times \frac{7 R}{2 \times 1.4} Δ T + PdV$
$= (\frac{70}{2 \times 14}) PdV + PdV$
$= \frac{7}{2} PdV$
$dQ = \frac{7}{2} \times 200$
$dQ = 700 J$

TS EAMCET 18.07.2022

Thermodynamics

148379 A monoatomic gas does $100 J$ of work when it is expanded isobarically. How much of heat is given to the gas in the process

1

150 J

2

200 J

3

250 J

4

300 J

Explanation:

C Given,
Gas is mono atomic
$∴ ∴ γ = \frac{5}{3}$
$C_{P} - C_{V} = R [Mayer's equation]$
$\frac{C_{P}}{C_{V}} = γ$
$C_{P} = γ C_{V}$
$C_{V} (γ - 1) = R$
$C_{V} = \frac{R}{γ - 1}$
Now from $1^{st}$ Law of thermodynamics
$dQ = dU + dW$
$= {mC}_{V} dT + PdV$
$= \frac{m \times RdT}{γ - 1} + PdV$
$dQ = \frac{PdV}{\frac{5}{3} - 1} + PdV$
$= \frac{3}{2} PdV + PdV$
$dQ = \frac{5}{2} \times PdV$
$dQ = \frac{5}{2} \times 100$
$dQ = 250 J$

TS EAMCET 18.07.2022

Thermodynamics

148380 If the temperature of $50 ml$ of gas at $27^{\circ} C$ raised to $57^{\circ} C$ at constant pressure, the final volume of the gas is

1

23.7 ml

2

25.0 ml

3

53.7 ml

4

55.0 ml

Explanation:

D Given,
Initial volume $(V_{1}) = 50 ml$
Initial temperature $(T_{1}) = 27^{\circ} C = 300 K$
Final Temperature $(T_{2}) = 57^{\circ} C = 330 K$
Ideal gas equation
$PV = nRT$
For constant pressure process
$\frac{V}{T} =$ Constant
$∴ \frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$
$\frac{50}{300} = \frac{V_{2}}{330}$
$V_{2} = 55 ml$

TS EAMCET 03.05.2018

Thermodynamics

148381 A thermodynamic system is taken from an original state $D$ to an intermediate state $E$ by linear process shown in the figure. Its volume is then reduced to the original volume from $E$ to $F$ by an isobaric process. The total work done by the gas from $D$ to $E$ to $F$ will be.

1

- 450 J

2

450 J

3

900 J

4

1350 J

Explanation:

B Given

We know that,
Area under the $P - V$ diagram is equal to work done.
$Δ W = \frac{1}{2} \times 300 \times 3 = 450 J$

JEE Main-29.07.2022

Thermodynamics

148378 A diatomic gas $(C_{p} = \frac{7}{2} R)$ does $200 J$ of work when it is expanded isobarically. The heat given to the gas in the process is

1

600 J

2

800 J

3

900 J

4

700 J

Explanation:

D Given,
$\frac{C_{P}}{C_{V}} = γ \Rightarrow C_{V} = \frac{C_{P}}{γ}$
$C_{P} = \frac{7}{2} R, For diatomic$
$C_{v} = \frac{7 R}{2 γ} = \frac{7 R}{2 \times 1.4}$
We know that,
From $1^{st}$ law of thermodynamics
$dQ = dU + dW$
$dQ = {mC}_{v} Δ T + dW$
$= m \times \frac{7 R}{2 \times 1.4} Δ T + PdV$
$= (\frac{70}{2 \times 14}) PdV + PdV$
$= \frac{7}{2} PdV$
$dQ = \frac{7}{2} \times 200$
$dQ = 700 J$

TS EAMCET 18.07.2022

Thermodynamics

148379 A monoatomic gas does $100 J$ of work when it is expanded isobarically. How much of heat is given to the gas in the process

1

150 J

2

200 J

3

250 J

4

300 J

Explanation:

C Given,
Gas is mono atomic
$∴ ∴ γ = \frac{5}{3}$
$C_{P} - C_{V} = R [Mayer's equation]$
$\frac{C_{P}}{C_{V}} = γ$
$C_{P} = γ C_{V}$
$C_{V} (γ - 1) = R$
$C_{V} = \frac{R}{γ - 1}$
Now from $1^{st}$ Law of thermodynamics
$dQ = dU + dW$
$= {mC}_{V} dT + PdV$
$= \frac{m \times RdT}{γ - 1} + PdV$
$dQ = \frac{PdV}{\frac{5}{3} - 1} + PdV$
$= \frac{3}{2} PdV + PdV$
$dQ = \frac{5}{2} \times PdV$
$dQ = \frac{5}{2} \times 100$
$dQ = 250 J$

TS EAMCET 18.07.2022

Thermodynamics

148380 If the temperature of $50 ml$ of gas at $27^{\circ} C$ raised to $57^{\circ} C$ at constant pressure, the final volume of the gas is

1

23.7 ml

2

25.0 ml

3

53.7 ml

4

55.0 ml

Explanation:

D Given,
Initial volume $(V_{1}) = 50 ml$
Initial temperature $(T_{1}) = 27^{\circ} C = 300 K$
Final Temperature $(T_{2}) = 57^{\circ} C = 330 K$
Ideal gas equation
$PV = nRT$
For constant pressure process
$\frac{V}{T} =$ Constant
$∴ \frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$
$\frac{50}{300} = \frac{V_{2}}{330}$
$V_{2} = 55 ml$

TS EAMCET 03.05.2018

Thermodynamics

148381 A thermodynamic system is taken from an original state $D$ to an intermediate state $E$ by linear process shown in the figure. Its volume is then reduced to the original volume from $E$ to $F$ by an isobaric process. The total work done by the gas from $D$ to $E$ to $F$ will be.

1

- 450 J

2

450 J

3

900 J

4

1350 J

Explanation:

B Given

We know that,
Area under the $P - V$ diagram is equal to work done.
$Δ W = \frac{1}{2} \times 300 \times 3 = 450 J$

JEE Main-29.07.2022