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A perfect gas goes from state A to anoth...

A perfect gas goes from state A to another state B by absorbing 8 x `10^5` J of heat and doing 6.5 x `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,

A

(a)Work done on gas is `10^5J`

B

(b)Work done on gas is `0.5 xx10^5J`

C

(c)Work done by gas is `10^5J`

D

(d)Work done by gas is `0.5 xx10^5J`

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To solve the problem step by step, we will use the first law of thermodynamics, which states that the change in internal energy (ΔU) of a system is equal to the heat added to the system (ΔQ) minus the work done by the system (ΔW). The equation can be expressed as: \[ \Delta Q = \Delta U + \Delta W \] ### Step 1: Analyze the first process (from state A to state B) Given: - Heat absorbed (ΔQ₁) = \(8 \times 10^5 \, \text{J}\) - Work done by the gas (ΔW₁) = \(6.5 \times 10^5 \, \text{J}\) Using the first law of thermodynamics: \[ \Delta U_1 = \Delta Q_1 - \Delta W_1 \] Substituting the values: \[ \Delta U_1 = (8 \times 10^5) - (6.5 \times 10^5) \] Calculating ΔU₁: \[ \Delta U_1 = 1.5 \times 10^5 \, \text{J} \] ### Step 2: Analyze the second process (between the same states) In the second process, we have: - Heat absorbed (ΔQ₂) = \(10^5 \, \text{J}\) We will use the same change in internal energy (ΔU) calculated from the first process, which is \(1.5 \times 10^5 \, \text{J}\). Using the first law of thermodynamics again for the second process: \[ \Delta Q_2 = \Delta U + \Delta W_2 \] Substituting the known values: \[ 10^5 = 1.5 \times 10^5 + \Delta W_2 \] ### Step 3: Solve for work done in the second process (ΔW₂) Rearranging the equation to find ΔW₂: \[ \Delta W_2 = 10^5 - 1.5 \times 10^5 \] Calculating ΔW₂: \[ \Delta W_2 = 10^5 - 1.5 \times 10^5 = -0.5 \times 10^5 \, \text{J} \] ### Conclusion The negative sign indicates that work is done on the gas during the second process. Therefore, the work done on the gas is: \[ \Delta W_2 = -0.5 \times 10^5 \, \text{J} \] ### Final Answer The work done on the gas in the second process is \(0.5 \times 10^5 \, \text{J}\).
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