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One mole of a monatomic gas is taken fro...

One mole of a monatomic gas is taken from a point `A` to another point `B` along the path `ACB`. The initial temperature at `A` is `T_(0)`. Calculate the heat abosrbed by the gas in the process `A rarr C rarr B`.

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The correct Answer is:
11
If `T_B` be the temperature at B, then by gas law
`(P_A V_A)/(T_B)=P_B/T_B" ":. T_B = (P_BV_B)/(P_A T_A)T_A=((2P_0)( 2V_0))/(P_0 V_0) T_0`
The change in internal energy from A to B
`Delta U = nC_v Delta T = 11 xx(3R)/2 xx(2V_0 - V_0)" "=(9RT_0)/2`
Work done in the process A to C `W_(AC) = P Delta V = P_0 (2V_0 - V_0)`
` = P_0 V_0 = R T_0` and `W_(CB) = 0" ":. " Total work done form " A to C to B `
`W_(AC) + Q_(CB) = RT_(0) + 0 = RT_0` From first law of thermodynamics
`Q = Detla U = W " "=(9 RT_0)/2 + RT_0 = (11 RT_0)/2`
Thus heat absorbed by the gas from ` A to C to B` is `(11 RT_0)/2`
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