at what temprature the root mean square velocity of a molecule of gas will be double of the value at 0 celcius?
回答 (2)
2017-11-06 6:24 pm
Root means square velocity, v(r.m.s.) = √(3RT/M), where :
R is gas constant ;
T is temperature in K ; and
M is molar mass of the gas.
v(r.m.s.) at T K = v(r.m.s.) at 0°C
√(3RT/M) = 2 ×√[3R(273)/M]
√T = 2 × √273
Temperature, T = 4 × 273 K = 1092 K = 819°C
R is gas constant ;
T is temperature in K ; and
M is molar mass of the gas.
v(r.m.s.) at T K = v(r.m.s.) at 0°C
√(3RT/M) = 2 ×√[3R(273)/M]
√T = 2 × √273
Temperature, T = 4 × 273 K = 1092 K = 819°C
2017-11-06 6:21 pm
RMSV is proportional to √T, so we would need to quadruple the temperature in order to double the RMSV.
4 x 273.15 K = 1092.6 K
4 x 273.15 K = 1092.6 K
收錄日期: 2021-04-18 17:56:57
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