Question

For a monoatomic gas, work done at constant pressure is W. The heat supplied at constant volume for the same rise in temperature of the gas is

• A. W
• B. $\frac{5W}{2}$
• C. $\frac{W}{2}$
• D. $\frac{3W}{2}$

Hint:

For any ideal gas, the ratio of work done at constant pressure to heat at constant volume is matching the degree factor: $\frac{W}{Q_v} = \frac{R}{C_v}$. Since $C_v = 1.5R$ for a monoatomic gas, the work is roughly $\frac{2}{3}$ of the internal energy, making $Q_v = 1.5W$ instantly!

Answer 1

The Correct Option is D

According to the first law of thermodynamics for a process at constant pressure, the work done ($W$) is: $$W = P\Delta V = nR\Delta T$$ The heat supplied at constant volume ($Q_v$) for the exact same temperature rise ($\Delta T$) is equal to the change in internal energy: $$Q_v = nC_v\Delta T$$ For an ideal monoatomic gas, the molar heat capacity at constant volume is $C_v = \frac{3}{2}R$. Substituting this into the internal energy equation: $$Q_v = n\left(\frac{3}{2}R\right)\Delta T = \frac{3}{2}(nR\Delta T)$$ Since $W = nR\Delta T$, we substitute it directly into our expression: $$Q_v = \frac{3W}{2}$$
Final Answer:
The heat supplied at constant volume is $\frac{3W}{2}$, which corresponds to option (D).