3.6gm of an ideal gas was injected into a bulb of internal volume of 8.221L at pressure Patm And temp Tk . The bulb was then placed in a themostat maintained at (T+15)k. .6 gm of the gas was let off keep the original pressure . find Pand T if mol weight of gas is 36

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Prakhar singh · Answer

T =75 k P=0.075 Use formula n1t1=n2t2 And get the value of t and then put it into PV=nRT N1=1/10 moles N2 = 1/12moles T1= T k T2= (T+15)k n1t1=n2t2 T=75k PV=nRT P= 0.0821*75/82.1= 0.075 atm

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3.6gm of an ideal gas was injected into a bulb of internal volume of 8.221L at pressure Patm And temp Tk . The bulb was then placed in a themostat maintained at (T+15)k. .6 gm of the gas was let off keep the original pressure . find Pand T if mol weight of gas is 36

3.6gm of an ideal gas was injected into a bulb of internal volume of 8.221L at pressure Patm And temp Tk . The bulb was then placed in a themostat maintained at (T+15)k. .6 gm of the gas was let off keep the original pressure . find Pand T if mol weight of gas is 36

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3.6gm of an ideal gas was injected into a bulb of internal volume of 8.221L at pressure Patm And temp Tk . The bulb was then placed in a themostat maintained at (T+15)k. .6 gm of the gas was let off keep the original pressure . find Pand T if mol weight of gas is 36

3.6gm of an ideal gas was injected into a bulb of internal volume of 8.221L at pressure Patm And temp Tk . The bulb was then placed in a themostat maintained at (T+15)k. .6 gm of the gas was let off keep the original pressure . find Pand T if mol weight of gas is 36

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