The internal energy (U) of an ideal gas is plotted against volume for a cyclic process A B C D A, as shown in the figure. The temperature of the gas at B and C are 500 K and 300 K, respectively, The heat absorbed by the gas (in cal/ mol) in this cyclic process, is:

Question

The internal energy (U) of an ideal gas is plotted against volume for a cyclic process A B C D A, as shown in the figure. <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/9665d284427fee84a1c690ff89b05cad-.png" /> The temperature of the gas at B and C are 500 K and 300 K, respectively, The heat absorbed by the gas (in cal/ mol) in this cyclic process, is: (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

q = q AB + q BC + q CD + q DA = nRT B ⋅ ln (2 V 0/ V 0)+ nC v , m ( T C - T B )+ nRT c ln ( V 0/2 V 0) + nC v , m ( T A - T D ) = nR ln 2( T B - T C ) =1 × 2 × 0.7 × 200=280 cal / mol

Q. The internal energy ( $U$ ) of an ideal gas is plotted against volume for a cyclic process $A BC D A$ , as shown in the figure.

The temperature of the gas at $B$ and $C$ are $500 K$ and $300 K$ , respectively, The heat absorbed by the gas (in cal/ mol) in this cyclic process, is:

$q = q_{A B} + q_{BC} + q_{C D} + q_{D A}$
$= n R T_{B} \cdot ln \frac{2 V _{0}}{V _{0}} + n C_{v, m} (T_{C} - T_{B}) + n R T_{c} ln \frac{V _{0}}{2 V _{0}}$
$+ n C_{v, m} (T_{A} - T_{D})$
$= n R ln 2 (T_{B} - T_{C})$
$= 1 \times 2 \times 0.7 \times 200 = 280 c a l / m o l$

Q. The internal energy (U) of an ideal gas is plotted against volume for a cyclic process ABCDA, as shown in the figure. The temperature of the gas at B and C are 500K and 300K, respectively, The heat absorbed by the gas (in cal/ mol) in this cyclic process, is:

q=qAB​+qBC​+qCD​+qDA​ =nRTB​⋅lnV0​2V0​​+nCv,m​(TC​−TB​)+nRTc​ln2V0​V0​​ +nCv,m​(TA​−TD​) =nRln2(TB​−TC​) =1×2×0.7×200=280cal/mol

Q. The internal energy ( $U$ ) of an ideal gas is plotted against volume for a cyclic process $A BC D A$ , as shown in the figure.

The temperature of the gas at $B$ and $C$ are $500 K$ and $300 K$ , respectively, The heat absorbed by the gas (in cal/ mol) in this cyclic process, is:

$q = q_{A B} + q_{BC} + q_{C D} + q_{D A}$
$= n R T_{B} \cdot ln \frac{2 V _{0}}{V _{0}} + n C_{v, m} (T_{C} - T_{B}) + n R T_{c} ln \frac{V _{0}}{2 V _{0}}$
$+ n C_{v, m} (T_{A} - T_{D})$
$= n R ln 2 (T_{B} - T_{C})$
$= 1 \times 2 \times 0.7 \times 200 = 280 c a l / m o l$