JEE Advance - Chemistry (1993 - No. 14)
A gas bulb of 1 litre capacity contains 2.0 $$\times$$ 1021 molecules of nitrogen exerting a pressure of 7.57 $$\times$$ 103 Nm-2. Calculate the root mean square (r.m.s) speed and the temparature of the gas molecules. If the ratio of the most probable speed to the root mean square speed is 0.82, calculate the most probable speed for these molecules at this temparature.
vrms = 494.2 m/s, T = 274.2 K, vmp = 405.2 m/s
vrms = 490 m/s, T = 270 K, vmp = 400 m/s
vrms = 500 m/s, T = 280 K, vmp = 410 m/s
vrms = 480 m/s, T = 260 K, vmp = 390 m/s
vrms = 510 m/s, T = 290 K, vmp = 420 m/s
Објашњење
According to gas equation,
$$PV = nRT$$ ; $$n = {{2 \times {{10}^{21}}} \over {6.02 \times {{10}^{23}}}}$$
$$T = {{PV} \over {nR}}$$
$$ = {{7.57 \times {{10}^3}\,N{m^{ - 2}} \times 1 \times {{10}^{ - 3}}\,{m^3}} \over {{{2 \times {{10}^{21}}} \over {6.02 \times {{10}^{23}}}}\,mol \times 8.314\,J\,(Nm)\,mo{l^{ - 1}}\,{K^{ - 1}}}}$$
$$ = 274.2$$ K
RMS velocity,
$$u = \sqrt {{{3RT} \over M}} = \sqrt {{{3 \times 8.314 \times 274.2} \over {28 \times {{10}^{ - 3}}}}} $$
$$ = 494.2$$ ms$$-$$1
Most probable velocity = 0.82 $$\times$$ u
= 494.2 $$\times$$ 0.82 ms$$-$$1 = 405.2 ms$$-$$1