Question

Figure below shows two paths that may be taken by a gas to go from a state A to a state C.
In process AB, 400J of heat is added to the system and in process BC, 100J of heat is added to the system. The heat absorbed by the system in the process AC will be?
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Answer 1

EXPLANATION.

In process AB, 400J of heat is added to the system.

In process BC, 100J of heat is added to the system.

As we know that,

Formula of :

⇒ dQ = dU + dW.

⇒ dW = ∫pdv.

As we know that,

AB is a isochoric process.

⇒ dV = 0 also dW = 0.

Heat supplied : $Q_{AB} = + 400 J$

Applying First law of thermodynamics.

⇒ ΔU = Q - W.

⇒ ΔU = 400J - 0.

⇒ ΔU = 400J. = $\Delta U_{AB} = + 400j$

In the process of BC.

⇒ W = ∫pdv.

⇒ $W = \int P (V_{C} - V_{B} )$

⇒ W = (6 x 10⁴) x [4 x 10⁻³ - 2 x 10⁻³].

⇒ W = (6 x 10⁴) x [2 x 10⁻³].

⇒ W = 6 x 2 x 10.

⇒ W = 120J.

⇒ $W_{BC} = 120J \ \ \ and \ \ \ Q_{BC} = 100J$

Applying First law of thermodynamics.

⇒ ΔU = Q - W.

⇒ ΔU = 100 - 120.

⇒ ΔU = - 20J.

⇒ $U_{BC} = - 20J$

Now, we can write equation as,

⇒ $U_{AB} + U_{BC} + U_{CA} = 0$

⇒ $400J + (-20J) + U_{CA} = 0$

⇒ $U_{CA} = - 380J. = U_{AC} = 380J.$

⇒ $Q_{AC} = U_{AC} + W_{AC}$

⇒ $W_{AC} = \dfrac{1}{2} (2 \times 10^{4} + 6 \times 10^{4} ) \times (2 \times 10^{-3} )$

⇒ $W_{AC} = 8 \times 10^{4} \times 10^{-3} = 80J$

⇒ $Q_{AC} = 380J + 80J = 460J$

Answer 2

Answer:

Qu€stion:-

Figure below shows two paths that may be taken by a gas to go from a state A to a state C.

In process AB, 400J of heat is added to the system and in process BC, 100J of heat is added to the system. The heat absorbed by the system in the process AC will be?

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AN$wer :-

In process AB, 400J of heat is added to the system.

In process BC, 100J of heat is added to the system.

As we know that,

Formula of :

⇒ dQ = dU + dW.

⇒ dW = ∫pdv.

As we know that,

AB is a isochoric process.

⇒ dV = 0 also dW = 0.

Heat supplied : $Q_{AB} = + 400 J$

Applying First law of thermodynamics.

⇒ ΔU = Q - W.

⇒ ΔU = 400J - 0.

⇒ ΔU = 400J. = $\Delta U_{AB} = + 400j$

In the process of BC.

⇒ W = ∫pdv.

⇒ $W = \int P (V_{C} - V_{B} )$

⇒ W = (6 x 10⁴) x [4 x 10⁻³ - 2 x 10⁻³].

⇒ W = (6 x 10⁴) x [2 x 10⁻³].

⇒ W = 6 x 2 x 10.

⇒ W = 120J.

⇒ $W_{BC} = 120J \ \ \ and \ \ \ Q_{BC} = 100J$

Applying First law of thermodynamics.

⇒ ΔU = Q - W.

⇒ ΔU = 100 - 120.

⇒ ΔU = - 20J.

⇒ $U_{BC} = - 20J$

Now, we can write equation as,

⇒ $U_{AB} + U_{BC} + U_{CA} = 0$

⇒ $400J + (-20J) + U_{CA} = 0$

⇒ $U_{CA} = - 380J. = U_{AC} = 380J.$

⇒ $Q_{AC} = U_{AC} + W_{AC}$

⇒ $W_{AC} = \dfrac{1}{2} (2 \times 10^{4} + 6 \times 10^{4} ) \times (2 \times 10^{-3} )$

⇒ $W_{AC} = 8 \times 10^{4} \times 10^{-3} = 80J$

⇒ $Q_{AC} = 380J + 80J = 460J$

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