Question

The First Law of Thermodynamics states that heat is a form of energy, and thermodynamic processes are therefore subject to the principle of conservation of energy. This means that heat energy cannot be created or destroyed. It can, however, be transferred from one location to another and converted to and from other forms of energy.

The equation for the first law of thermodynamics is given as;

$\Delta U=q+W$

Where,

One mole of an ideal gas $C_v=\frac{5}{2}R\text{at}300\text{K}\text{and}5\text{atm}$ is expanded adiabatically to a final pressure of $2\text{atm}$ against a constant pressure of $2\text{atm}$. Find the temperature of the gas?

• A. $270\text{K}$
• B. $273\text{K}$
• C. $248.5\text{K}$
• D. $200\text{K}$

Answer 1

The correct option is C $248.5\text{K}$

Given:
$P_{ext}=2\text{atm}$
$P_1=5\text{atm}$
$P_2=2\text{atm}$
$T_1=300\text{K}$
$T_2=?$
$n=1$
$C_v=\frac{5}{2}R$
For an adiabatic process:
$W=n{C}_{v}dT=-P\Delta V$
$n{C}_v(T_2-T_1)=-P(V_2-V_1)$
$\Rightarrow 1\times \frac{5}{2}R(T_2-300)=-P(\frac{nR{T}_2}{P_2}-\frac{nR{T}_1}{P_1})$
$\Rightarrow \frac{5}{2}R(T_2-300)=-2R(\frac{T_2}{2}-\frac{300}{5})$
$\Rightarrow \frac{5}{2}(T_2-300)=-2(\frac{T_2}{2}-60)$
$\Rightarrow 5T_2-1500=-2T_2+240$
$\Rightarrow 7T_2=1740$
$\Rightarrow T_2=248.57\text{K}$