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Question: To what pressure must a certain ideal gas \(\left( {\gamma = 1.4} \right)\) at \(373K\),\(1\) atmosp...

To what pressure must a certain ideal gas (γ=1.4)\left( {\gamma = 1.4} \right) at 373K373K,11 atmospheric pressure be compressed adiabatically in order to raise its temperature to 773K773K?
(A). 7.89 atm7.89{\text{ atm}}
(B). 79 atm79{\text{ atm}}
(C). 7.1 atm 7.1{\text{ atm }}
(D). 6.89 atm6.89{\text{ atm}}

Explanation

Solution

As Know the question the gas is compressed adiabatically so we will make use of equation for adiabatic process that is,
PVγ=P{V^\gamma } = Constant. The adiabatic gas equation holds true for only ideal gases.

Complete step by step answer:
The term “adiabatic” refers to the elements that prevent heat transfer with the environment. So, the adiabatic process is the one in which there is no exchange of heat from the system to the surroundings neither during compression nor during expansion.Some examples of adiabatic process are:
1. The release of air from a pneumatic tire.
2. When we put ice into an ice box, no heat comes in and no heat goes out.
The following conditions are followed under this process.
The system must be perfectly insulated from surroundings.
The process must be carried out quickly so that there is enough time for heat transfer to take place.
The adiabatic process equation is :
PVr= constantP{V^r} = {\text{ constant}}
Adiabatic compression of any gas is defined as the compression in which no heat is added or subtracted from the and ΔV\Delta V (internal energy ) is increased that is equal to the external work done. The pressure of the gas is more than the volume as tempertin tends to decrease during compression.
We are given with γ=1.4\gamma = 1.4
So P1γTγ={P^{1 - \gamma }}{T^\gamma } = constant
p1{p_1} \to Initial pressure
p1= 1 atm{{\text{p}}_1} = {\text{ 1 atm}}
P2{{\text{P}}_2} \to Final pressure and let p2= p{{\text{p}}_2} = {\text{ p}}
T1373K{{\text{T}}_1} \to 373K and T2773K{{\text{T}}_2} \to 773K
So, calculating,
(p1p2)1γ=(T2T1)γ{\left( {\dfrac{{{p_1}}}{{{p_2}}}} \right)^{1 - \gamma }} = {\left( {\dfrac{{{T_2}}}{{{T_1}}}} \right)^\gamma }
Substituting,
(1p)11.4=(773373)1.4{\left( {\dfrac{1}{p}} \right)^{1 - 1.4}} = {\left( {\dfrac{{773}}{{373}}} \right)^{1.4}}
(1p)0.4=(2.07)1.4{\left( {\dfrac{1}{p}} \right)^{ - 0.4}} = {\left( {2.07} \right)^{1.4}}
(p)0.4=(2.07)1.4{\left( p \right)^{0.4}} = {\left( {2.07} \right)^{1.4}}
On solving, p=7.89 atmp = 7.89{\text{ atm}}
So, option A is correct.

Note: The adiabatic process also includes expansion of gas. The adiabatic expansion is defined as an ideal behaviour of a closed system in which the pressure is constant and the temperature remains decreasing.