Question

300 J of energy is given to a gas at constant volume which increases its temperature from $20^\circ$C to $50^\circ$C. If $R = 8.3$ S.I. units and $C_v = \dfrac{5R}{2}$, find mass of gas.

Hint:

In isochoric processes, supplied heat only changes internal energy since work done is zero.

Answer 1

Correct Answer: 4

Step 1: Identify thermodynamic process.
Since volume is constant, the process is isochoric.
Step 2: Heat equation for isochoric process.
\[ Q = n C_v \Delta T \]
Step 3: Substituting given values.
\[ 300 = n \cdot \dfrac{5R}{2} (50 - 20) \]
Step 4: Solving for number of moles.
\[ 300 = n \cdot \dfrac{5R}{2} \cdot 30 \Rightarrow n = \dfrac{4}{R} \]
Step 5: Mass of gas.
\[ \text{Mass of gas} = \dfrac{4}{R} \times \text{molecular weight} \]
(Note: Molecular weight is not given.)