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Question: A vessel containing \(10\,litre\) of air under a pressure of \(1\,MPa\) is connected to a \(4\,litre...

A vessel containing 10litre10\,litre of air under a pressure of 1MPa1\,MPa is connected to a 4litre4\,litre empty vessel. What is the final air pressure in the vessel assuming that the process is isothermal?

Explanation

Solution

Hint
Here we have to apply the concept of the ideal gas equation.
The ideal gas law is the equation of the state of a hypothetical ideal gas, sometimes called the general gas equation. Under many conditions, it is a good approximation of the behaviour of many gases, though it has several limitations. Benoît Paul Émile Clapeyron first stated this in 18341834
As a combination of the empirical law of Boyle, the law of Charles, the law of Avogadro and the law of Gay-Lussac. The ideal gas law is often written mathematically as-
PV=nRTPV = nRT
Where PP, VV and TT are the pressure, volume and temperature respectively, nn is the amount of substance and RR is the ideal gas constant.

Complete step by step answer
Given,First volume, V1=10litres{V_1} = 10\,litres
Second volume, V2=4litre{V_2} = 4\,litre
First pressure, P1=1MPa{P_1} = 1\,MPa
Second pressure, P2=?{P_2} = ?
Also, the process is isothermal.
An isothermal method is a thermodynamic process in which a system’s temperature stays constant. Heat transfer into or out of the system occurs so slowly that it maintains thermal equilibrium.
An isothermal process is a thermodynamic method in which a system’s temperature stays constant. Heat transfer into or out of the system takes place so slowly that it maintains thermal equilibrium.
So, here the temperature is constant
So, nPVn \propto PV
Here as nn is the amount of substance, so, it will remain conserved along with the product PVPV .
Hence,
P2(V1+V2)=P1V1\Rightarrow {P_2}\left( {{V_1} + {V_2}} \right) = {P_1}{V_1}
P2(10+4)=1×10\Rightarrow {P_2}\left( {10 + 4} \right) = 1 \times 10
P2=57MPa\Rightarrow {P_2} = \dfrac{5}{7}\,MPa
Hence, the final air pressure in the vessel, assuming that the process is isothermal is 57MPa \dfrac{5}{7}\,MPa.

Note
Here we have to see whether the process is isothermal or adiabatic or isotropic. If the process would have been given as something else other than isothermal, then the temperature would not have remained constant. So, we have to pay attention to the question.