SATHEE: Thermodynamics Question 142

Thermodynamics Question 142

Question: A perfect gas goes from state A to another state B by absorbing $8 \times 10^{5} J$ of heat and doing $6.5 \times 10^{5} J$ of external work. It is now transferred between the same two states in another process in which it absorbs $10^{5} J$ of heat. Then in the second process

[BHU 1997]

Options:

A) Work done on the gas is $0.5 \times 10^{5} J$

B) Work done by gas is $0.5 \times 10^{5} J$

C) Work done on gas is $10^{5} J$

D) Work done by gas is $10^{5} J$

Show Answer

Answer:

Correct Answer: A

Solution:

In first process using $Δ Q = Δ U + Δ W$

Therefore $8 \times 10^{5} = Δ U + 6.5 \times 10^{5}$

Therefore $Δ U = 1.5 \times 10 J$

Since final and initial states are same in both process

So $Δ U$ will be same in both process

For second process using $Δ Q = Δ U + Δ W$

Therefore $10^{5} = 1.5 \times 10^{5} + Δ W$

Therefore $Δ W = - 0.5 \times 10^{5} J$

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Thermodynamics Question 142

Question: A perfect gas goes from state A to another state B by absorbing 8×105J of heat and doing 6.5×105J of external work. It is now transferred between the same two states in another process in which it absorbs 105J of heat. Then in the second process

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Question: A perfect gas goes from state A to another state B by absorbing $8 \times 10^{5} J$ of heat and doing $6.5 \times 10^{5} J$ of external work. It is now transferred between the same two states in another process in which it absorbs $10^{5} J$ of heat. Then in the second process