An ideal gas follows a process described by the equation $PV^2 = C$ from the initial $(P_1, V_1, T_1)$ to final $(P_2, V_2, T_2)$ thermodynamic states,where $C$ is a constant. Then:

  • A
    If $P_1 > P_2$ then $T_1 < T_2$
  • B
    If $V_2 > V_1$ then $T_2 > T_1$
  • C
    If $V_2 > V_1$ then $T_2 < T_1$
  • D
    If $P_1 > P_2$ then $V_1 > V_2$

Explore More

Similar Questions

$A$ gas expands with temperature according to the relation $V = k T^{2/3}$. What is the work done when the temperature changes by $30^{\circ}C$ (in $R$)?

The volume of $1 \; mole$ of an ideal gas with the adiabatic exponent $\gamma$ is changed according to the relation $V = \frac{b}{T}$,where $b$ is a constant. The amount of heat absorbed by the gas in the process if the temperature is increased by $\Delta T$ will be:

$0.02 \, mol$ of an ideal diatomic gas with initial temperature $20^{\circ} C$ is compressed from $1500 \, cm^3$ to $500 \, cm^3$. The thermodynamic process is such that $p V^2 = \beta$,where $\beta$ is a constant. Then,the value of $\beta$ is close to (the gas constant,$R = 8.31 \, J / K / mol$).

Work done to increase the temperature of one mole of an ideal gas by $30^{\circ} C$, if it is expanding under the condition $V \propto T^{2/3}$ is, $(R = 8.314 \ J/mol \cdot K)$ (in $J$)

Hydrogen gas is undergoing a process given by $PV^2 = \text{constant}$. The ratio of work done by the gas to the change in its internal energy is

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo