Question

According to the given graph match the correct statements, given below line gives the thermodynamics process and α is volume coefficient and T is the temperature of the ideal gas

• A. If θ = 45°, then the given process is isobaric process
• B. If θ = 37°, then the pressure, volume and temperature of the gas are increasing from point (1) to point (2)
• C. If θ = 53°, then the equation of the process is T4V–3 = constant

Answer 1

Answer: (A)

If θ = 45°, then the given process is isobaric process

Answer 2

Key idea: On a T–V graph for an ideal gas, an isobaric process satisfies

$$\frac{T}{V}=\text{constant}\quad\Longrightarrow\quad \frac{dT}{dV}=1\quad(\theta=45^\circ).$$

• For θ=45°, $\tan\theta=1\implies \frac{dT}{dV}=1\implies T/V=\text{constant}\implies P=\frac{RT}{V}=\text{constant}$.
• For θ=37°, $\tan37^\circ=3/4$, so $T/V\ne\text{constant}$ → not isobaric and $P$ does not increase.
• For θ=53°, $\tan53^\circ=4/3$ corresponds to a different polytropic exponent, not $T\,V^{-3}=\text{constant}.$