A poly-atomic molecule ( = 3R, = 4R, and R is gas constant) goes from phase space point A () to point B () to point C (). A to B is an adiabatic path and B to C is an isothermal path. The net heat absorbed per unit mole by the system is:. [202

Question

A poly-atomic molecule ( $C_{V}$ = 3R, $C_{P}$ = 4R, where R is gas constant) goes from phase space point A ( $P_{A} = 10^{5} Pa, V_{A} = 4 \times 10^{- 6} m^{3}$ ) to point B ( $P_{B} = 5 \times 10^{4} Pa, V_{B} = 6 \times 10^{- 6} m^{3}$ ) to point C ( $P_{C} = 10^{4} Pa, V_{C} = 8 \times 10^{- 6} m^{3}$ ). A to B is an adiabatic path and B to C is an isothermal path.

The net heat absorbed per unit mole by the system is:. [2025]

Smart Achievers · Accepted Answer

(2)

For process A $\to$ B

${(△ Q)}_{A B} = 0$ (adiabatic)

For process B $\to$ C

${(△ Q)}_{B C} = {(△ W)}_{B C} = n R T \ln (\frac{V_{C}}{V_{B}})$

= 450 R [ln (4) – ln 3]

${(△ Q)}_{net} = {(△ Q)}_{A B} + {(△ Q)}_{B C}$

= 450 R [ln (4) – ln (3)]

Solution

1 500 R (ln 3 + ln 4)

2 450 R (ln 4 – ln 3)

3 500 R ln 2

4 400 R ln 4

Ans.
(2)

For process A $\to$ B

${(△ Q)}_{A B} = 0$ (adiabatic)

For process B $\to$ C

${(△ Q)}_{B C} = {(△ W)}_{B C} = n R T \ln (\frac{V_{C}}{V_{B}})$

= 450 R [ln (4) – ln 3]

${(△ Q)}_{net} = {(△ Q)}_{A B} + {(△ Q)}_{B C}$

= 450 R [ln (4) – ln (3)]

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