Question

The value of $\Delta H$ for cooling 2 mole of an ideal monoatomic gas from ${225}^{o}C$ to ${125}^{o}C$ at constant pressure will be $(given{C}_p=\frac{5}{2}R)$

• A. 250 R
• B. -500 R
• C. 500 R
• D. -250 R

Answer 1

The correct option is B -500 R

Given,

$n=2$ mole, $T_1={225}^{o}C=498K$ and $T_2={125}^{o}C=398K$ ,

$C_p=5/2R$

Since enthalpy change is given as

$\Delta H=n{C}_p\Delta T$

$=n{C}_p(T_2-T_1)$

$=2\times \frac{5}{2}R(498-398)$

$\Delta H=5R\times -100$

$\Delta H=-500R$

Hence, the correct option is $\text{B}$