3.6 grams of the ideal gas in bulb of volume 8 lit and the pressure is p atmp the temperature TK is placed in the thermostat at the temperature T+15k of 0.6 geams of gads at original pressure find the PAndT of gas if the weight is 44
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P1 = n1.R.T1/V1
P1 = n2.R.T1 + 50/V1
n1.T1 = n2.T1 + 50
T1. n1/n2 - 1= 50
n1/n2 = m1/m2
T1.m1/m2 - 1= 50
T1.4.0 g / 3. 2 g - 1= 50
T1.1.25 - 1.00= 50
T1 = 50/ 0.25= 200 K
T = T1 = 200 K
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Initial moles, n=
3.6/44
Final moles, n '
=
3.6−0.6/44
=
3.0/44
Since P and V are constant,
n 1t1=n2t2
method 2...
in attachment
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