SATHEE: Thermodynamics Question 271

Thermodynamics Question 271

Question: The variation of internal energy (U) with the pressure (P) of 2.0 mole gas in cyclic process abcda. The temperature of gas at c and d are 300 K and 500 K. calculate the heat absorbed by the gas during the process.

Options:

A) 400 R In 2

B) 200 R In 2

C) 100 R In 2

D) 300 R In 2

Show Answer

Answer:

Correct Answer: A

Solution:

[a] Change in internal energy for cyclic process $(Δ U) = 0.$

For process $a \to b$ , (P-constant) $W_{a \to b} = P Δ V$

$= n R Δ T = - 400 R$

For process $b \to c$ (T-constant) $W_{b \to c} = - 2 R (300) ln 2$

For process $c \to d$ , (P-constant) $W_{c \to d} = + 400 R$

For process $d \to a$ , (T-constant) $W_{d \to a} = + 2 R (500) ln 2$

Net work (AW) $= W_{a \to b} + W_{b \to c} + W_{c \to d} + W_{d \to a}$

$Δ W = 400 R ln 2$

$∴ d Q = d U + d W$ , first law of thermodynamics $∴ d Q = 400 R ln 2.$

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Thermodynamics Question 271

Question: The variation of internal energy (U) with the pressure (P) of 2.0 mole gas in cyclic process abcda. The temperature of gas at c and d are 300 K and 500 K. calculate the heat absorbed by the gas during the process.

Options:

Answer:

Solution:

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