Two cylinders contain the same amount of an ideal monatomic gas. The same amount of heat is given to both cylinders. If the temperature rise in cylinder $A$ is $T_0$,then the temperature rise in cylinder $B$ will be:

  • A
    $\frac{4}{3} T_0$
  • B
    $2 T_0$
  • C
    $\frac{T_0}{2}$
  • D
    $\frac{5}{3} T_0$

Explore More

Similar Questions

$A$ diatomic ideal gas undergoes a thermodynamic change according to the $P-V$ diagram shown in the figure. The heat given to the gas in process $AB$ is: (in $, P_0 V_0$)

In an isobaric process of an ideal gas, the ratio of work done by the system $(W)$ during the expansion and the heat exchanged $(Q)$ is $(\gamma = \frac{C_p}{C_v})$

The ratio of specific heats of a gas is $\gamma$. The change in internal energy of one mole of the gas, when the volume changes from $V$ to $2V$ at constant pressure $p$, is:

One mole of an ideal gas is contained within a cylinder by a frictionless piston and is initially at temperature $T$. The pressure of the gas is kept constant while it is heated and its volume doubles. If $R$ is the molar gas constant,the work done by the gas in increasing its volume is

Which of the following relations is correct for an isochoric process?

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