Question

1 mole of gas expands isothermally at 37°C. The amount of heat is absorbed by it until its volume doubled is (R = 8.31 J mol–1 K–1)

• A. 411.25 cal
• B. 418.50 cal
• C. 420.25 cal
• D. 425.40 cal

Answer 1

Answer: (D)

425.40 cal

Answer 2

: here,

$V _ { 2 } = 2 V _ { 1 }$

Work done in isothermal process is

$W = n R T \ln \frac { V _ { 2 } } { V _ { 1 } } = 1 \times 8.31 \times 310 \times \ln \frac { 2 V _ { 1 } } { V _ { 1 } }$

$= 8.31 \times 310 \times \ln 2 = 1.786 \times 10 ^ { 3 }$

Amount of heat absorbed

$= \frac { 1.786 \times 10 ^ { 3 } } { 4.2 } \mathrm { cal } = 425.4 \mathrm { cal }$