An ideal monoatomic gas is carried along the cycle A B C D A as shown in the figure. The total heat absorbed during this process is

Question

An ideal monoatomic gas is carried along the cycle A B C D A as shown in the figure. The total heat absorbed during this process is (A) 10.5 p 0 V 0 (B) 7.5 p 0 V 0 (C) 2.5 p0 V0 (D) 1.5 p 0 V 0

Tardigrade Admin · Accepted Answer

Heat absorbed means heat is extracted from source, i.e.

Q

must be positive.
This occurs along path

A \to B \to C

∴

Heat added

= Δ Q_{A BC} = Δ U_{A BC} + Δ W_{A BC}

= n C_{V} (T_{C} - T_{A}) + 3 p_{0} (2 V_{0} - V_{0})

= n \frac{3}{2} R (T_{C} - T_{A}) + 3 p_{0}

V_{0} = \frac{3}{2} (n R T_{C} - n R T_{A}) + 3 p_{0} V_{0}

= \frac{3}{2} (p_{C} V_{C} - p_{A} V_{A}) + 3 p_{0} V_{0}

= \frac{3}{2} (3 p_{0} 2 V_{0} - p_{0} V_{0}) + 3 p_{0} V_{0}

= \frac{3}{2} \times 5 p_{0} V_{0} + 3 p_{0}

V_{0} = \frac{21}{2} p_{0} V_{0} = 10.5 p_{0} V_{0}

Q. An ideal monoatomic gas is carried along the cycle $A BC D A$ as shown in the figure. The total heat absorbed during this process is

$10.5 p_{0} V_{0}$

$7.5 p_{0} V_{0}$

$2.5 p_{0} V_{0}$

$1.5 p_{0} V_{0}$

Q. An ideal monoatomic gas is carried along the cycle ABCDA as shown in the figure. The total heat absorbed during this process is

10.5p0​V0​

7.5p0​V0​

2.5p0​V0​

1.5p0​V0​

Q. An ideal monoatomic gas is carried along the cycle $A BC D A$ as shown in the figure. The total heat absorbed during this process is

$10.5 p_{0} V_{0}$

$7.5 p_{0} V_{0}$

$2.5 p_{0} V_{0}$

$1.5 p_{0} V_{0}$