Answer to Question #270323 in Molecular Physics | Thermodynamics for Anu

Question

A gas has a volume of 120m³ at stp. if the gas is compressed adiabatically to a volume of 25m³. calculate the new temperature, pressure and work done on the gas. [Cp= 1.68kj/kg-¹ k-¹, Cv = 1.20 kj kg-¹ k-¹]

Answer 1

For adiabatic compression

$P_{2}=P_{1}(\frac{V_{1}}{V_{2}} )^r$

Where:

$P_{1}=$ initial pressure= atmospheric pressure$=1.00×10^5N/M^{2}$

$P_{2}=$ final pressure

$V_{1}=$ Initial volume=$120M^{3}$

$V_{2}=$ final volume=$25M^3$

$r=\frac{Cv}{Cp}$

$r=$ $\frac{1.68kj/kg^{-1}K^{-1}}{1.2kj/kg^{-1}K^{-1}} =1.4$

Therefore:

$P_{2}=(1×10^5N/M^2)×(\frac{120M^3}{25M^3})^{1.4}=8.99×10^5N/M^2$

From Ideal gas laws Final Temperature:

$T_{2}=(\frac{P_{2}V_{2}}{P_{1}V_{1}})T_{1}$

Where:

$T_{1}= Room temperature=293K$

$T_{2}= (\frac{8.99×10^5N/M^2×25M^3}{1×10^5N/M^2×120M^3})×293K=548.76K$

Work done in Adiabatic process W:

$W=\frac{1}{1-r}(P_{2} V_{2}-P_{1} V_{1})$

$W=\frac{1}{1-1.4}(8.99×10^5N/M^2×25M^3-1×10^5N/M^2×120M^3)= -261.875j$