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Calculate the work done when one mole of an ideal monoatomic gas is compressed adiabatically. The initial pressure and volume of the gas are 10^5N//m^2 and 6 litre respectively. The final volume of the gas is 2litres. Molar specific heat of the gas at constant volume is 3R/2. |
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Answer» Solution :For an adiabatic change `P_1V_1^(GAMMA)=P_2V_2^(gamma)` or `P_2=P_1(V_1/V_2)^(gamma)` Here `P_1=10^5 N//m^2, V_1` =6 litre , `V_2` =2 litreand for a monoatomicgas , we have `C_V=(3R)/2` and `C_P=(3R)/2+R=(5R)/2` or `gamma=C_P/C_V=(5R//2)/(3R//2)=5/3` Thus `P_2=10^5xx(6/2)^(5//3)=10^5xx(3)^(5//3)` We know that work done on gas `DeltaW`in adiabatic change is given by `DeltaW=(P_2V_2-P_1V_1)/(gamma-1)` `=(10^5xx(3)^(5//3)xx(2xx10^(-3))-10^5 xx(6xx10^(-3)))/((5//3)-1)` [Here `V_2`=2 litre `=2xx10^(-3)m^3`and `V_1`=6 litre= `6xx10^(-3)m^3` ] `=(2xx10^2[(3)^(5//3)-3])/(2//3)` `=(2xx10^2[6.19-3])/(2//3)=3xx10^2xx(3.19)` =957 J |
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