In the process shown in figure- the internal energy of an ideal gas decreases by 3 P 0 V 0-2 in going from point C to A- Heat transfer along the process CA isA- ZeroB- -3 P 0 V 0-2C- -5 P 02 V 0-2D- -3 P 0 V 0

The correct option is C 5P_0 V_0/2 Process C → A : – Δ Q=Δ U+Δ W Δ U= 3P_0 V_0/2 Δ W=area under curve =P_0 (V_0 2V_0) = P_0 V_0 ∴Δ Q= 3P_0 V_0/2 P_0 V_0 ...

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Standard XII

Chemistry

Internal Energy

In the proces...

Question

In the process shown in figure, the internal energy of an ideal gas decreases by $\frac{3 P_{0} V_{0}}{2}$ in going from point C to A. Heat transfer along the process CA is

Zero

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$\frac{- 3 P_{0} V_{0}}{2}$

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$\frac{- 5 P_{0} V_{0}}{2}$

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$- 3 P_{0} V_{0}$

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Solution

The correct option is C
$\frac{- 5 P_{0} V_{0}}{2}$

Process $C \to A$ : –

$Δ Q = Δ U + Δ W$

$Δ U = \frac{- 3 P_{0} V_{0}}{2}$

$Δ$ W=area under curve

$= P_{0} (V_{0} - 2 V_{0})$

$= - P_{0} V_{0}$

$∴ Δ Q = \frac{- 3 P_{0} V_{0}}{2} - P_{0} V_{0}$

$= \frac{- 5 P_{0} V_{0}}{2}$

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In the process shown in figure, the internal energy of an ideal gas decreases by 3P0V02 in going from point C to A. Heat transfer along the process CA is

The correct option is C −5P0V02 Process C→A : – ΔQ=ΔU+ΔW ΔU=−3P0V02 Δ W=area under curve =P0(V0−2V0) =−P0V0 ∴ΔQ=−3P0V02−P0V0 =−5P0V02

In the process shown in figure, the internal energy of an ideal gas decreases by $\frac{3 P_{0} V_{0}}{2}$ in going from point C to A. Heat transfer along the process CA is

The correct option is C
$\frac{- 5 P_{0} V_{0}}{2}$

Process $C \to A$ : –

$Δ Q = Δ U + Δ W$

$Δ U = \frac{- 3 P_{0} V_{0}}{2}$

$Δ$ W=area under curve

$= P_{0} (V_{0} - 2 V_{0})$

$= - P_{0} V_{0}$

$∴ Δ Q = \frac{- 3 P_{0} V_{0}}{2} - P_{0} V_{0}$

$= \frac{- 5 P_{0} V_{0}}{2}$