One mole of a monatomic ideal gas (R = 8.314 J mol^-1K^-1) undergoes a cycle: 1->2 isochoric heating from 300 K until the pressure doubles; 2->3 reversible isothermal expansion until the pressure returns to its initial value; 3->1 isobaric compression to the initial state. Which statements are correct?

The heat absorbed in step 1->2 is about 3741 J.

The net heat absorbed in the cycle equals the net work done by the gas, about 963 J.

The work done on the gas in step 3->1 is about 2494 J.

The heat absorbed in step 1->2 is about 3741 J.

The net heat absorbed in the cycle equals the net work done by the gas, about 963 J.

The work done on the gas in step 3->1 is about 2494 J.

The heat released in step 3->1 is about 3741 J.

JEE Advanced levelOne or more correct optionsDifficulty 3 of 3td015

One mole of a monatomic ideal gas ( $R = 8.314 J mo l^{- 1} K^{- 1}$ ) undergoes a cycle: $1 \to 2$ isochoric heating from $300 K$ until the pressure doubles; $2 \to 3$ reversible isothermal expansion until the pressure returns to its initial value; $3 \to 1$ isobaric compression to the initial state. Which statements are correct?

AThe heat absorbed in step $1 \to 2$ is about $3741 J$ .
BThe net heat absorbed in the cycle equals the net work done by the gas, about $963 J$ .
CThe work done on the gas in step $3 \to 1$ is about $2494 J$ .
DThe heat released in step $3 \to 1$ is about $3741 J$ .

Step 1:

q = C_{V} Δ T = 1.5 R (300) = 3741.3 J

(true). Step 2 occurs at

600 K

with

V_{3} / V_{2} = 2

q_{23} = RT ln 2 = 3457.7 J

. Step 3:

V

halves at constant

P

T

falls

600 \to 300

, so

q = C_{P} Δ T = - 2.5 R (300) = - 6235.5 J

and

w = + P Δ V

magnitude

R (300) = 2494.2 J

(true). Net

q = 963.5 J > 0

and

Δ U = 0

so net work done by gas equals it (true). The heat released in step 3 is

6236 J

, not

3741 J

(false).