Question

When \( Q \) amount of heat is supplied to a monatomic gas, the work done by the gas is \( W \). When \( Q_1 \) amount of heat is supplied to a diatomic gas, the work done by the gas is \( 2W \). Then \( Q:Q_1 \) is: 

• A. \( 2:3 \)
• B. \( 3:5 \)
• C. \( 5:7 \)
• D. \( 5:14 \)

Hint:

For ideal gases, the heat supplied in an isoic process is given by \( Q = \Delta U + W \). The values of \( \Delta U \) and \( W \) depend on the degrees of freedom of the gas.

Answer 1

The Correct Option is D

Step 1: Understanding the Relationship Between Heat and Work 
For an ideal gas undergoing an isoic process, the first law of thermodynamics states: \[ Q = \Delta U + W. \] For a monatomic gas: \[ \Delta U = \frac{3}{2} nR \Delta T, \quad W = nR \Delta T. \] Thus, the heat supplied: \[ Q = \Delta U + W = \frac{3}{2} nR \Delta T + nR \Delta T = \frac{5}{2} nR \Delta T. \] For a diatomic gas: \[ \Delta U = \frac{5}{2} nR \Delta T, \quad W = 2nR \Delta T. \] Thus, the heat supplied: \[ Q_1 = \Delta U + W = \frac{5}{2} nR \Delta T + 2nR \Delta T = \frac{9}{2} nR \Delta T. \] 

Step 2: Ratio Calculation 
\[ \frac{Q}{Q_1} = \frac{\frac{5}{2} nR \Delta T}{\frac{9}{2} nR \Delta T} = \frac{5}{14}. \] 

Step 3: Conclusion 
Thus, the ratio is: \[ \boxed{5:14}. \]