Start Free Mock LoginAlready registered? Login here

Physics - Heat & Thermo Question with Solution | TestHub

PhysicsHeat & ThermoThermodynamics Laws & Specific Heats of GasesEasy2 minQB

PhysicsEasynumerical

Two identical containers A and B at the same temperature and volume $V$ hold masses $m_{1}$ and $m_{2}$ of the same ideal gas. Both expand isothermally to $3 V$ . The pressure change in B is 1.25 times that in A. Find $\frac{m _{1}}{m _{2}}$ .

Answer:

0.80

Solution:

Isothermal: $P_{f} = \frac{P _{i}}{3}$ , so the fall in pressure is $\frac{2}{3} P_{i} \propto m$ . $Δ P_{B} = 1.25Δ P_{A}$ gives $m_{2} = 1.25 m_{1}$ , so $\frac{m _{1}}{m _{2}} = 0.8$ .

Subject:PhysicsTopic:Heat & ThermoSubtopic:Thermodynamics Laws & Specific Heats of Gases

⏱ 2mℹ️ Source: QB

Doubts & Discussion

Loading discussions...