A certain amount of gas of volume V at 27°C temperature and pressure 2 × 107 Nm–2 expands isothermally until its volume gets doubled. Later it expands adiabatically until its volume gets redoubled. The final pressure of the gas will be (Use, γ = 1.5)
\(3.536\times10^5 \)Pa
\(3.536\times10^6\) Pa
\(1.25\times10^6\) Pa
\(1.25\times10^5 \)Pa
The Correct Option is B
Solution and Explanation
P1=2×107 Pa
P1V1=P2V2
Since V2=2V1 Hence P2=\(\frac{P_1}{2}\) (isothermal expansion)
P2=1×107Pa
P2V2γ=P2V2γ
P3=\(\frac{1×10^7}{2^1.5}\)=3.536×106
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Concepts Used:
Ideal Gas Equation
An ideal gas is a theoretical gas composed of a set of randomly-moving point particles that interact only through elastic collisions.
What is Ideal Gas Law?
The ideal gas law states that the product of the pressure and the volume of one gram molecule of an ideal gas is equal to the product of the absolute temperature of the gas and the universal gas constant.
PV=nRT
where,
P is the pressure
V is the volume
n is the amount of substance
R is the ideal gas constant
Ideal Gas Law Units
When we use the gas constant R = 8.31 J/K.mol, then we have to plug in the pressure P in the units of pascals Pa, volume in the units of m3 and the temperature T in the units of kelvin K.