Question

1 mole of ideal monoatomic gas is taken along a cyclic process as shown in figure.Process 1 to2 shown in one by 4 th part of circle as shown dotted line process 2 to 3 is isochoric while 3 to 1 is isobaric. Of efficiency of cycle to the nearest integer isn%value of n is
​

Answer 1

Heat rejected = 5PoVo/ R

Q = 3PoVo

Explanation:

Heat rejected by the gas in CA ΔQ = nCpΔT, where n = 1

Cp = 5R/2 for monoatomic gas

PV = nRT

Ti = Po * 2Vo /R

Tf = PoVo/ R

ΔT = PoVo/ R- 2PoVo/ R

Substituting the values, we get:

Q = 5R/2 * (-PoVo/ R) = -5PoVo/ R

Negative sign indicates that heat is given out.

So heat rejected = 5PoVo/ R

Heat absorbed in AB is ΔQ = nCpΔT

Cv = 3R/2 for monoatomic gas

Ti = PoVo/ R

Tf = 3PoVo/ R

So Q = 3R/2 * (3PoVo/ R - PoVo/ R)

    Q = 3PoVo