One mole of an ideal monoatomic gas is caused to go through the cycle shown in the figure. Then the change in the internal energy in expanding the gas from $a$ to $c$ along the path $a b c$ is:

3 P_{0} V_{0}

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6 R T_{0}

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4.5 R T_{0}

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10.5 R T_{0}

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Solution

The correct option is C $3 P_{0} V_{0}$
$a ⟶ b$ is a isobaric process.
$∴ W_{a b} = P_{o} (Δ V) = P_{o} (4 V_{o} - V_{o}) = 3 P_{o} V_{o}$
$b ⟶ c$ is a isobaric process.
$∴ Δ V = 0 ∴ W_{b c} = 0 ∴ Δ U = W_{a b} + W_{b c} = 3 P_{o} V_{o} + 0 = 3 P_{o} V_{o}$

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One mole of an ideal monoatomic gas is caused to go through the cycle shown in the figure. Then the change in the internal energy in expanding the gas from a to c along the path abc is:

The correct option is C 3P0V0a⟶b is a isobaric process. ∴Wab=Po(ΔV)=Po(4Vo−Vo)=3PoVo b⟶c is a isobaric process.∴ΔV=0∴Wbc=0∴ΔU=Wab+Wbc=3PoVo+0=3PoVo

One mole of an ideal monoatomic gas is caused to go through the cycle shown in the figure. Then the change in the internal energy in expanding the gas from $a$ to $c$ along the path $a b c$ is:

The correct option is C $3 P_{0} V_{0}$
$a ⟶ b$ is a isobaric process.
$∴ W_{a b} = P_{o} (Δ V) = P_{o} (4 V_{o} - V_{o}) = 3 P_{o} V_{o}$
$b ⟶ c$ is a isobaric process.
$∴ Δ V = 0 ∴ W_{b c} = 0 ∴ Δ U = W_{a b} + W_{b c} = 3 P_{o} V_{o} + 0 = 3 P_{o} V_{o}$