Volume versus temperature graph of two moles of helium gas is as shown in figure. The ratio of heat absorbed and the work done by the gas in process 1-2 is

Question

Volume versus temperature graph of two moles of helium gas is as shown in figure. The ratio of heat absorbed and the work done by the gas in process 1-2 is <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-pkbsxkzlv7ic6ipr.jpg" /> (A) 3 (B) 5/2 (C) 5/3 (D) 7/2

Tardigrade Admin · Accepted Answer

V-T graph is a straight line passing through origin. Hence, V propto T or P= constant (Δ Q/Δ W)=(1/1 - 3 / 5)=(5/2)

Q. Volume versus temperature graph of two moles of helium gas is as shown in figure. The ratio of heat absorbed and the work done by the gas in process $1 - 2$ is

$3$

$5/2$

$5/3$

$7/2$

$V - T$ graph is a straight line passing through origin. Hence, $V \propto T$ or $P =$ constant $\frac{Δ Q}{Δ W} = \frac{1}{1 - 3/5} = \frac{5}{2}$

Q. Volume versus temperature graph of two moles of helium gas is as shown in figure. The ratio of heat absorbed and the work done by the gas in process 1−2 is

3

5/2

5/3

7/2

V−T graph is a straight line passing through origin. Hence, V∝T or P= constant ΔWΔQ​=1−3/51​=25​

Q. Volume versus temperature graph of two moles of helium gas is as shown in figure. The ratio of heat absorbed and the work done by the gas in process $1 - 2$ is

$3$

$5/2$

$5/3$

$7/2$

$V - T$ graph is a straight line passing through origin. Hence, $V \propto T$ or $P =$ constant $\frac{Δ Q}{Δ W} = \frac{1}{1 - 3/5} = \frac{5}{2}$