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Thermal Physics
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19 Mar 2010 02:17:58 IST
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use the formula q= ncdt.......now here process is carried out at constant volume so ...q=nc (v) dt
...that is 1 . 2 . 3R ....put the value of r u will get the answer......here n=2.......dt=1............and for a polyatomic gas c(v) is 3R!! .......R=8.314 J/mol K













Moles of Methane = n = 2
Molar specific heat of methane at constant volume, Cv = 3R J/K/mol
Thus heat required to heat methane by 1 deg C is = nCv dT = 2*3R*1 = 6R J = 6 * 8.314 J =49.9 J
or heat =49.9/4.2 cal = 11.88 Cal