Question

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

Answer 1

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,
$\mathrm{\Delta }U=q+W$
$W=\mathrm{\Delta }U-q$ ---(i)
Here ΔU is the change in the internal energy system and W is the work done by the system.
We know
$W=P\mathrm{\Delta }V$----(ii)
From (i) & (ii)
$P\mathrm{\Delta }V=\mathrm{\Delta }U-q$
$\mathrm{\Delta }U=q+P\mathrm{\Delta }V$---(iii)

Also, we know
$\mathrm{\Delta }U=n{C}_V\mathrm{\Delta }T$
Initial internal energy of n moles of diatomic gas is
$U_1=n\left({\displaystyle \frac{6}{2}}R\right)T_1$=$3nR{T}_1$
Now, Internal energy of gas after heating is
$U_2=n\left({\displaystyle \frac{6}{2}}R\right)T_2$=$3nR{T}_2$
Therefore,
$\mathrm{\Delta }U=U_2-U_1$
$\mathrm{\Delta }U=3nR{T}_2-3nR{T}_1$
$\mathrm{\Delta }U=3nR(T_2-T_1)$
$\mathrm{\Delta }U=3nR\mathrm{\Delta }T$

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

From (iii) and (iv)
$3P\mathrm{\Delta }V=q+P\mathrm{\Delta }V$
$2P\mathrm{\Delta }V=q$
$P\mathrm{\Delta }V={\displaystyle \frac{q}{2}}$
$W={\displaystyle \frac{800}{2}}$
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