Question

2m³ volume of a gas at a pressure of 4 × 10⁵ Nm⁻² is compressed adiabatically so that its volume becomes 0.5m³. Find the new pressure. Compare this with the pressure that would result if the compression was isothermal. Calculate work done in each process. γ = 1.4.


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Answer 1

$\large\underline{\underline{\mathbb{\color{magenta}{GIVEN :}}}}$

2m³ volume of a gas at a pressure of 4 × 10⁵ Nm⁻² is compressed adiabatically so that its volume becomes 0.5m³.

• V₁ = 2m³
• P₁ =  4 × 10⁵ Nm⁻²
• V₂ = 0.5m³
• γ = 1.4

$\large\underline{\underline{\mathbb{\color{magenta}{TO\ FIND :}}}}$

• The new pressure.
• Compare this with the pressure that would result if the compression was isothermal.
• Calculate work done in each process.

$\large\underline{\underline{\mathbb{\color{magenta}{SOLUTION :}}}}$

In adiabatic process,

$\sf{P_1V_1^{\gamma}=P_2V_2^{\gamma}}$

$\sf{\implies P_2=4\times 10^5 \bigg[\dfrac{2}{0.5} \bigg]^{1.4}}$

$\sf{\implies P_2=4\times10^5(4)^{1.4}}$

$\boxed{\sf{\implies P_2=2.8\times10^6\ Nm^{-2}}}$

____________________

In isothermal process,

$\sf{P_1V_1=P_2V_2}$

$\sf{\implies P_2= \dfrac{P_1V_1}{V_2} }$

$\sf{\implies P_2= \dfrac{4\times10^5\times2}{0.5} }$

$\boxed{\sf{\implies P_2=1.6\times10^6\ Nm^{-2} }}$

♛________________________♛

Now,

Work done in adiabatic process -

$\sf{W=\dfrac{P_2V_2-P_1V_1}{\gamma-1} }$

$\sf{\implies W=\dfrac{(2.8 \times 10^6 \times 0.5)-(4\times10^5\times2)}{1.4-1} }$

$\boxed{\sf{\implies W=1.48\times10^6\ J }}$

____________________

Work done in isothermal process -

$\sf{W = 2.3026RT\ log \dfrac{V_2}{V_1}}$

$\sf{\implies W=2.3026P_1V_1\ log\dfrac{V_2}{V_1}}$

$\sf{\implies W= 2.3026\times4\times10^5\times 2log \bigg(\dfrac{1}{4} \bigg) }$

$\boxed{\sf{\implies W= -1.1\times10^2\ J}}$

Answer 2

$\huge{\red{\boxed{\underline{\underline{\mathtt{\color{gold}{Given:↓}}}}}}}$

2m³ volume of a gas at a pressure of 4 × 10⁵ Nm⁻² is compressed adiabatically so that its volume becomes 0.5m³.

V₁ = 2m³

V₁ = 2m³P₁ = 4 × 10⁵ Nm⁻²

V₁ = 2m³P₁ = 4 × 10⁵ Nm⁻²V₂ = 0.5m³

V₁ = 2m³P₁ = 4 × 10⁵ Nm⁻²V₂ = 0.5m³γ = 1.4

$\huge{\red{\boxed{\underline{\underline{\mathtt{\color{magenta}{To\:Find:↓}}}}}}}$

The new pressure.

Compare this with the pressure that would result if the compression was isothermal.

Calculate work done in each process.

$\huge{\red{\boxed{\underline{\underline{\mathtt{\color{navy}{Solution:↓}}}}}}}$

In adiabatic process,

$\sf{P_1V_1^{\gamma}=P_2V_2^{\gamma}}$

$\sf{\implies P_2=4\times 10^5 \bigg[\dfrac{2}{0.5} \bigg]^{1.4}}$

$\sf{\implies P_2=4\times10^5(4)^{1.4}}$

$\boxed{\sf{\implies P_2=2.8\times10^6\ Nm^{-2}}}$

____________________

In isothermal process,

$\sf{P_1V_1=P_2V_2}P$

$\sf{\implies P_2= \dfrac{P_1V_1}{V_2} }$

$\sf{\implies P_2= \dfrac{4\times10^5\times2}{0.5} }$

$\boxed{\sf{\implies P_2=1.6\times10^6\ Nm^{-2} }}$

♛________________________♛

Now,

Work done in adiabatic process -

$\sf{W=\dfrac{P_2V_2-P_1V_1}{\gamma-1} }$

$\sf{\implies W=\dfrac{(2.8 \times 10^6 \times 0.5)-(4\times10^5\times2)}{1.4-1} }$

$\boxed{\sf{\implies W=1.48\times10^6\ J }}$

____________________

Work done in isothermal process -

$\sf{W = 2.3026RT\ log \dfrac{V_2}{V_1}}$

$\sf{\implies W=2.3026P_1V_1\ log\dfrac{V_2}{V_1}}$

$\sf{\implies W= 2.3026\times4\times10^5\times 2log \bigg(\dfrac{1}{4} \bigg) }$

$\boxed{\sf{\implies W= -1.1\times10^2\ J}}$