1g h2 gas expand at stp to occupy double of its original volume
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The question is incomplete, so I am assuming that it means to calculate the work done.
Since the condition given is at STP conditions, we can use the formula W=nRT ln(v2/v1) to calculate work done.
Since the expansion takes place in isothermal process n = 0.5 gm moles
Initial Vol is v and upon expansion at STP conditions, it becomes doubled to 2v. The temperature is kept constant.
W=nRT ln(v2/v1)
=0.5 x 8.314 x 273 x ln(2v/v)
=786.6257 joules
Upon conversion to calories, W = 786.6257 / 4.184 calories
Therefore, work done W = 187.8 calories
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