Question

The ratio of the work done, change in internal energy and heat absorbed when a diatomic gas expands at constant pressure is

• A. 2:3:5
• B. 7:5:2
• C. 5:3:2
• D. 2:5:7

Hint:

For any ideal gas process, the ratio \(W : \Delta U : Q_p\) in an isobaric expansion is always \( (C_p - C_v) : C_v : C_p \). For a diatomic gas, this is \(R : \frac{5}{2}R : \frac{7}{2}R\), which simplifies to \(2:5:7\).

Answer 1

The Correct Option is D

Let's analyze the three quantities for an isobaric (constant pressure) process for n moles of a diatomic gas undergoing a temperature change \(\Delta T\).
1. Heat absorbed at constant pressure (\(Q_p\)):
\( Q_p = n C_p \Delta T \).
For a diatomic gas, the molar specific heat at constant pressure is \( C_p = \frac{7}{2}R \).
So, \( Q_p = n (\frac{7}{2}R) \Delta T \).
2. Change in internal energy (\(\Delta U\)):
The change in internal energy depends only on the temperature change: \( \Delta U = n C_v \Delta T \).
For a diatomic gas, the molar specific heat at constant volume is \( C_v = \frac{5}{2}R \).
So, \( \Delta U = n (\frac{5}{2}R) \Delta T \).
3. Work done (W):
From the first law of thermodynamics, \( Q = \Delta U + W \).
So, \( W = Q_p - \Delta U \).
\( W = n (\frac{7}{2}R) \Delta T - n (\frac{5}{2}R) \Delta T = n (\frac{2}{2}R) \Delta T = nR\Delta T \).
Alternatively, for a constant pressure process, \( W = P\Delta V \). From the ideal gas law \(PV=nRT\), we have \(P\Delta V = nR\Delta T\).
Now, we find the ratio \( W : \Delta U : Q_p \).
\( nR\Delta T : n (\frac{5}{2}R) \Delta T : n (\frac{7}{2}R) \Delta T \).
We can cancel the common factor \( nR\Delta T \).
The ratio is \( 1 : \frac{5}{2} : \frac{7}{2} \).
To express this with integers, we multiply all parts by 2.
The ratio becomes \( 2 : 5 : 7 \).