SATHEE: Heat and Thermodynamics 5 Question 5

Heat and Thermodynamics 5 Question 5

5. Following figure shows two processes $A$ and $B$ for a gas. If $Δ Q_{A}$ and $Δ Q_{B}$ are the amount of heat absorbed by the system in two cases, and $Δ U_{A}$ and $Δ U_{B}$ are changes in internal energies respectively, then

(2019 Main, 9 April I)

(a) $Δ Q_{A} > Δ Q_{B}, Δ U_{A} > Δ U_{B}$

(b) $Δ Q_{A} < Δ Q_{B}, Δ U_{A} < Δ U_{B}$

(d) $Δ Q_{A} = Δ Q_{B}; Δ U_{A} = Δ U_{B}$

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Answer:

Correct Answer: 5. (c)

Solution:

According to the first law of thermodynamics,

Heat supplied $(Δ Q) =$ Work done $(W) +$ Change in internal energy of the system $(Δ U)$

Similarly, for process $B$ ,

$Δ Q_{A} = Δ U_{A} + W_{A}$

Now, we know that,

$Δ Q_{B} = Δ U_{B} + W_{B}$

work done for a process $=$ area under it’s $p$ - $V$ curve Here,

Thus, it is clear from the above graphs,

$W_{A} > W_{B} \dots (i)$

Also, since the initial and final state are same in both process, so

$Δ U_{A} = Δ U_{B} \dots (i i)$

So, from Eqs. (i) and (ii), we can conclude that

$Δ Q_{A} > Δ Q_{B}$

Heat and Thermodynamics 5 Question 5

5. Following figure shows two processes $A$ and $B$ for a gas. If $Δ Q_{A}$ and $Δ Q_{B}$ are the amount of heat absorbed by the system in two cases, and $Δ U_{A}$ and $Δ U_{B}$ are changes in internal energies respectively, then

Answer:

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Heat and Thermodynamics 5 Question 5

5. Following figure shows two processes A and B for a gas. If ΔQA and ΔQB are the amount of heat absorbed by the system in two cases, and ΔUA and ΔUB are changes in internal energies respectively, then

Answer:

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5. Following figure shows two processes $A$ and $B$ for a gas. If $Δ Q_{A}$ and $Δ Q_{B}$ are the amount of heat absorbed by the system in two cases, and $Δ U_{A}$ and $Δ U_{B}$ are changes in internal energies respectively, then