The volume (V) of a monatomic gas varies with its temperature (T), as shown in the graph. The ratio of work done by the gas, to the heat absorbed by it, when it undergoes a change from state A to state B, is

Question

The volume (V) of a monatomic gas varies with its temperature (T), as shown in the graph. The ratio of work done by the gas, to the heat absorbed by it, when it undergoes a change from state A to state B, is (A) (2/7) (B) (2/5) (C) (1/3) (D) (2/3)

Tardigrade Admin · Accepted Answer

Given process is isobaric dQ = n Cp dT dQ = n ((5/2) R) dT dW = P dV = n RdT Required ratio = (dW/dQ) = (nRdT/n((5)2R)dT) = (2/5)

Q. The volume $(V)$ of a monatomic gas varies with its temperature $(T)$ , as shown in the graph. The ratio of work done by the gas, to the heat absorbed by it, when it undergoes a change from state $A$ to state $B$ , is

$\frac{2}{7}$

$\frac{2}{5}$

$\frac{1}{3}$

$\frac{2}{3}$

Given process is isobaric
$d Q = n C_{p} d T$
$d Q = n (\frac{5}{2} R) d T$
$d W = P d V = n R d T$
Required ratio $= \frac{d W}{d Q} = \frac{n R d T}{n ( \frac{5}{2} R ) d T} = \frac{2}{5}$

Q. The volume (V) of a monatomic gas varies with its temperature (T), as shown in the graph. The ratio of work done by the gas, to the heat absorbed by it, when it undergoes a change from state A to state B, is

72​

52​

31​

32​