Correct answer is (A)

In thermodynamics, the pressure-temperature graph for an ideal gas kept at constant volume (isochoric process) is a straight line. This is because for an ideal gas, the pressure is proportional to the temperature (as described by the ideal gas law, PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature).

It is clear from graph that for all the gases lines of graphs meet at same value.

At $\mathrm{x}$-axis (temperature axis) $\mathrm{P}$ is zero but temperature is negative and it will be equal to 0 K or $-{273}^{\circ }\mathrm{C}$.

Correct answer is (D)

$[a]=\left[{\mathrm{ML}}^5{\mathrm{T}}^{-2}\right]$

$\begin{array}{rl}& [b]=\left[{\mathrm{L}}^3\right]\\ \\ & \left[\frac{b^2}{a}\right]=\left[\frac{{\mathrm{L}}^6}{{\mathrm{ML}}^5{\mathrm{T}}^2}\right]=\left[{\mathrm{M}}^{-1}{\mathrm{LT}}^{-2}\right]\\ \\ & =[\text{ Compressibility] }\end{array}$

Correct answer is (B)

For isothermal process P-V graph is rectangular hyperbola

As the dotted line is an isobaric line which implies T₃ > T₂ > T₁ as volume is increasing.

Correct answer is (D)

Here volume is constant.

$\therefore$ $\frac{P_1}{T_1}=\frac{P_2}{T_2}$

$\Rightarrow \frac{100}{273}=\frac{180}{T_2}$

$\Rightarrow T_2=\frac{180}{100}\times 273$

$=1.8\times 273$

$=491K$

Correct answer is (C)

Total number of moles are

$n=n_{H_2}+n_{O_2}$

$=\frac{16}{2}+\frac{128}{32}$

= 12 moles

Using $PV=nRT$

$V=\frac{nRT}{P}$

$=\frac{12\times 8.31\times 273.15}{{10}^5}$ m³

= 0.27 m³ = 27 $\times$ 10⁴ cm³