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)

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

Tardigrade Admin · Accepted Answer

Here, T= 37 ° C = 37 + 273 = 310 K, V2 = 2V1 work done in isothermal process is W =nRT ln (V2/V1) = 1× 8.31 × 310 × ln (2V1/V1) =8.31 × 310 × ln 2 = 1.786 × 103 Amount of heat absorbed = (1.786 × 103/4.2) cal = 425.4 cal

Q. $1$ mole of gas expands isothermally at $3 7^{\circ} C$ . The amount of heat is absorbed by it until its volume doubled is $(R = 8.31 J m o l^{- 1} K^{- 1})$

$411.25$ cal

$418.50$ cal

$420.25$ cal

$425.40$ cal

Q. 1 mole of gas expands isothermally at 37∘C . The amount of heat is absorbed by it until its volume doubled is (R=8.31Jmol−1K−1)

411.25 cal

418.50 cal

420.25 cal

425.40 cal

Here, T=37∘C=37+273=310K,V2​=2V1​ work done in isothermal process is W=nRTlnV1​V2​​ =1×8.31×310×lnV1​2V1​​ =8.31×310×ln2=1.786×103 Amount of heat absorbed =4.21.786×103​ cal =425.4 cal

Q. $1$ mole of gas expands isothermally at $3 7^{\circ} C$ . The amount of heat is absorbed by it until its volume doubled is $(R = 8.31 J m o l^{- 1} K^{- 1})$

$411.25$ cal

$418.50$ cal

$420.25$ cal

$425.40$ cal