2 moles of an ideal monatomic gas is carried from a state (P0, V0) to a state (2P0, 2V0) along a straight line path in a P-V diagram. The amount of heat absorbed by the gas in the process is given by

Question

2 moles of an ideal monatomic gas is carried from a state (P0, V0) to a state (2P0, 2V0) along a straight line path in a P-V diagram. The amount of heat absorbed by the gas in the process is given by (A) 3P0 V0 (B) (9/2)P0 V0 (C) 6P0 V0 (D) (3/2)P0 V0

Tardigrade Admin · Accepted Answer

The internal energy Δ U=n Cv Δ T Cv= Specific heat of gas at constant volume ⇒ Δ U =n ⋅ (3 R/2)((4 p0 V0/n R)-(p0 V0/n R)) =n ⋅ (3 R/2)((4 p0 V0-p0 V0/n R)) =n ⋅ (3 R/2) ⋅ (3 p0 V0/n R) =(9/2) p0 V0 ldots(i) Work done by the gas W=(2 p0+p0) (V0/2)=(3 p0 V0/2) ldots(ii) From first law of thermodynamics, Δ Q =d W+d U =(3 p0 V0/2)+(9/2) p0 V0 [from Eqs. (i) and (ii)] =(3 p0 V0/2)+(9/2) p0 V0 =(12 p0 V0/2)=6 p0 V0

Q. $2$ moles of an ideal monatomic gas is carried from a state $(P_{0}, V_{0})$ to a state $(2 P_{0}, 2 V_{0})$ along a straight line path in a P-V diagram. The amount of heat absorbed by the gas in the process is given by

$3 P_{0} V_{0}$

$\frac{9}{2} P_{0} V_{0}$

$6 P_{0} V_{0}$

$\frac{3}{2} P_{0} V_{0}$

Q. 2 moles of an ideal monatomic gas is carried from a state (P0​,V0​) to a state (2P0​,2V0​) along a straight line path in a P-V diagram. The amount of heat absorbed by the gas in the process is given by

3P0​V0​

29​P0​V0​

6P0​V0​

23​P0​V0​

Q. $2$ moles of an ideal monatomic gas is carried from a state $(P_{0}, V_{0})$ to a state $(2 P_{0}, 2 V_{0})$ along a straight line path in a P-V diagram. The amount of heat absorbed by the gas in the process is given by

$3 P_{0} V_{0}$

$\frac{9}{2} P_{0} V_{0}$

$6 P_{0} V_{0}$

$\frac{3}{2} P_{0} V_{0}$