To an ideal trigonometric gas 800 Cal heat energy is given at constant pressure. If the vibrational mode is neglected, then energy used by gas in work done against surrounding is:
A. 200 Cal
B. 300 Cal
C. 400 Cal
D. 60 Cal

Hint: - We must know the ideal gas equation for solving this question. Hence our ideal gas equation is PV = nRT. It can be established from the postulates of the kinetic theory of gases developed by Clerk Maxwell.

Complete answer:
According to the first law of thermodynamics,
$\Delta U = q + W$
$W = \Delta U - q$ ---(i)
Here ΔU is the change in the internal energy system and W is the work done by the system.
We know
$W = P\Delta V$----(ii)
From (i) & (ii)
$P\Delta V = \Delta U - q$
$\Delta U = q + P\Delta V$---(iii)

Also, we know
$\Delta U = n{C_V}\Delta T$
Initial internal energy of n moles of diatomic gas is
${U_1} = n\left( {\dfrac{6}{2}R} \right){T_1}$=$3nR{T_1}$
Now, Internal energy of gas after heating is
${U_2} = n\left( {\dfrac{6}{2}R} \right){T_2}$=$3nR{T_2}$
Therefore,
$\Delta U = U_2 - U_1$
$\Delta U = 3nR{T_2} - 3nR{T_1}$
$\Delta U = 3nR({T_2} - {T_1})$
$\Delta U = 3nR\Delta T$

We know that in ideal gas equation,
$PV = nRT$
$P\Delta V = nR\Delta T$
$\Delta U = 3P\Delta V$---(iv)

From (iii) and (iv)
$3P\Delta V = q + P\Delta V$
$2P\Delta V = q$
$P\Delta V = \dfrac{q}{2}$
$W = \dfrac{{800}}{2}$
W=400 Cal from (i)

Hence option C is correct i.e., work done is 400 Cal

Note: - An ideal gas is a theoretical gas composed of many randomly moving point particles that are not subjected to interparticle interactions. The ideal gas concept is useful because it obeys the ideal gas law, a simplified equation of state, and it is also amenable to analysis under statistical mechanics