Question

$1{\text{m}}^3$ of helium originally at $0^{\circ }\text{C}$ and $1\text{atm}$ pressure is cooled at constant pressure until the volume becomes $0.75{\text{m}}^3$. How much heat has been removed from the system?
$(1\text{atm}={10}^5{\text{N/m}}^2,R=\frac{25}{3}\text{J/mol K})$

• A. $15\text{kcal}$
• B. $15\text{kJ}$
• C. $18\text{cal}$
• D. $18\text{kJ}$

Answer 1

The correct option is A $15\text{kcal}$

The equation of state of a gas during isobaric compression is given by
$V\propto T$
From this, we can say that,
$\frac{V_1}{T_1}=\frac{V_2}{T_2}$
From the data given in the question, we get
$T_2=\frac{0.75}{1}\times 273=204.75\text{K}$
From ideal gas equation, we know that,
$n=\frac{PV}{RT}......\left(1\right)$

Since Helium is monoatomic gas, $C_P$ for Helium is given as $C_P=\frac{5}{2}R.......\left(2\right)$
Heat given/taken during an isobaric process is given by
$\Delta Q=n{C}_P\Delta T$
From $\left(1\right)$ and $\left(2\right)$,
$\Delta Q=\frac{PV}{RT}\times \frac{5}{2}R(T_2-T_1)$
From the data given in the question,
$\Delta Q=\frac{{10}^5\times 1}{273}\times \frac{5}{2}(205-273)=-0.623\times {10}^5\text{J}$
$\Rightarrow \Delta Q=-0.148\times {10}^5\text{cal}\approx -15\text{kcal}$
Heat that is removed is $15\text{kcal}$.
Thus, option (a) is the correct answer.