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Physics 43 Online
OpenStudy (ujjwal):

Two moles of helium gas is contained in a container at STP. The heat energy needed to double the pressure of gas at constant volume is a)5052 J b)6005 J c)6800 J d)9200 J

OpenStudy (ujjwal):

where are the replies?

OpenStudy (anonymous):

i deleted them...dat was a wrong approach....was thinking of a new one.

OpenStudy (ujjwal):

ok

OpenStudy (ujjwal):

Energy of 1 mole Helium (monoatomic) gas= (3/2) RT So, Energy of 2 mole Helium gas at STP= (3/2) \(\times\)8.31 \(\times\) 273\(\times\)2 when pressure is doubled keeping volume constant, temperature is doubled! and So, The new value of energy after the pressure is doubled = \(\frac{3}{2}\)\(\times\)8.31\(\times\)273\(\times\)2\(\times\)2 So, difference in energy or energy required = 6805.89 J =6800 J (approx.) Thanks goes to @experimentX ..

OpenStudy (anonymous):

ah yes! of course! i was looking at the wrong place.... thanks @experimentX

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