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First Law of Thermodynamics Physics Thermodynamics Physics Class 11th

In a thermodynamic process, pressure of fixed mass of a gas is changed in such a manner that the gas releases 20 J of heat and 8 J of work is done on the gas. If initial internal energy of the gas was 40 J, what will be final internal energy?

Option 1 -

18 J

Option 2 -

28 J

Option 3 -

52 J

Option 4 -

32 J

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Answered by

alok kumar singh | Contributor-Level 10

9 hours ago

Correct Option - 2

Detailed Solution:

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An ideal gas undergoes the process $1 \to 2$ as shown in the figure, the heat supplied and work done in the process is $∆ Q$ and $∆ W$ respectively. The ratio $∆ Q : ∆ W$ is

alok kumar singh

From first law of thermodynamics

$? W = ? Q - ? U$

$? Q = n C_{p} ? T$

$? U = n C_{v} ? T$

$? W = n (C_{p} - C_{v}) ? T$

$\frac{? Q}{? W} = \frac{γ}{γ - 1} [\frac{C_{p}}{C_{p}} = γ]$

An ideal gas undergoes the process $1 \to 2$ as shown in the figure, the heat supplied and work done in the process is $∆ Q$ and $∆ W$ respectively. The ratio $∆ Q : ∆ W$ is

alok kumar singh

From first law of thermodynamics

$? W = ? Q - ? U$

$? Q = n C_{p} ? T$

$? U = n C_{v} ? T$

$? W = n (C_{p} - C_{v}) ? T$

$\frac{? Q}{? W} = \frac{γ}{γ - 1} [\frac{C_{p}}{C_{p}} = γ]$

1 mole of rigid diatomic gas performs a work of $\frac{Q}{5}$ when heat Q is supplied to it. The molar heat capacity of the gas during this transformation is The value of x is $\frac{x R}{8} .$ ………

[R = universal gas constant]

alok kumar singh

$Δ Q = Δ U + Δ W$

$\Rightarrow Q = Δ U + \frac{Q}{5} \Rightarrow Δ U = \frac{4 Q}{5} = n C_{v} Δ T \Rightarrow \frac{4 Q}{5} = \frac{5 R}{2} Δ T \Rightarrow Δ T = \frac{8 Q}{25 R}$

$Q = n c Δ T = 1 \times C \times \frac{80}{25 R} \Rightarrow C = \frac{25 R}{8} \Rightarrow x = 25$

A sample of an ideal gas is taken through the cyclic process ABCA as shown in figure. It absorbs, 40 J of heat during the part AB, no heat during BC and rejects 60 J of heat during CA. A work of 50 J is done on the during the part BC. The internal energy of the gas at A is 1560 J. The work done by the gas during the part CA is:

alok kumar singh

Given

Q_AB = +40J

Q_BC = 0J

Q_CA = -60J

W_CA =?

W_BC = -50J (on the gas)

U_A = 1560J

1^st Law on path A to B

W_AB + $Δ U = Q$

Þ 0 + $[U_{B} - U_{A}] = 40$

Þ U_B = U_A + 40

= 1560 + 40

U_B = 1600 J

1^st Low

...more

Given

Q_AB = +40J

Q_BC = 0J

Q_CA = -60J

W_CA =?

W_BC = -50J (on the gas)

U_A = 1560J

1^st Law on path A to B

W_AB + $Δ U = Q$

Þ 0 + $[U_{B} - U_{A}] = 40$

Þ U_B = U_A + 40

= 1560 + 40

U_B = 1600 J

1^st Low on path B to C

Q_BC = $Δ u + w_{B C}$

Þ 0 = U_c – U_D – 50

U_C = U_B + 50

= 1600 + 50

= 1650 J

U_A – U_C = $Δ U$ (Path C to A) = 1560-1650

= -90 J

1st Low for path C to A

Q_CA = $Δ U_{C A} + w_{C A} + w_{C A}$

Þ -60 = -90 + w_CA

W_CA = +30J

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In a thermodynamic process, pressure of fixed mass of a gas is changed in such a manner that the gas releases 20 J of heat and 8 J of work is done on the gas. If initial internal energy of the gas was 40 J, what will be final internal energy?

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