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First Law of Thermodynamics Physics Thermodynamics Physics Class 11th

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

Option 1 -

$γ : (γ - 1)$

Option 2 -

$γ$

Option 3 -

$(γ - 1)$

Option 4 -

$(γ - 1) : γ$

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Answered by

alok kumar singh | Contributor-Level 10

a month ago

Correct Option - 1

Detailed Solution:

From first law of thermodynamics

$? W = ? Q - ? U$

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

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

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

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

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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?

alok kumar singh

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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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

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An ideal gas undergoes the process 1 → 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

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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