11Someone Please Help
1)The specific gravity of CCl4 vapour at 273K and 76 cmHg in grams/litres ?
2)A monoatomic gas, a diatomic gas and a triatomic gas are mixed, taking one mole of each, the Cp/Cv for the mixture is ?
3)One mole of an ideal monoatomic gas is mixed with 1 mole of an ideal diatomic gas. The molar specific heat of the mixture at constant volume is ?
4)If the product of the gas constant R,i.e, 0.0821 lit atm/K/mole and NTP temperature in kelvin equals 22.4, the compressibility factor of the gas at 1 atm pressure is......... what temperature we shall take here, since they are telling NTP where T is 273K or 22.4K. am confused
5)The rms speed of N2 molecules in a gas sample can be changed by.
a)an increase in volume of the sample
b)mixing with Ar sample.
c)an increase in pressure on the gas
d)an increase in temperature.
in answers only option (d) is given correct, but rms speed is also directly proportional to pressure as well as volume, so won't it be changed by increasing the pressure & volume.
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5 Answers
1sol 1-
PV=(W/M)*RT
SO SPECIFIC GRAVITY =PM/RT....
SOL4- TAKE 273K
SOL5-IF U INCREASE PRESSURE THERE WILL BE DECREASE IN VOLUME..SO NET EFFECT OF CHNAGE IN VOLUME AND PRESSURE WILL BE ZERO...SO ONLY TEMPREATURE WILL EFFECT IT
11Ans 2)
Gamma = Cp/ Cv ---- (1). Gamma = 5/3 ( for monoatomic gas) , 7/5 (for diatomic gas) , 8/6 (for triatomic gas).
We also know from thermodnamics that Cp = Cv + R. ------- (2)
Using (1) amd (2) you can calculate Cp and Cv for each gas.
For monoatomic gas,
Cp/ Cv = 5/3 and Cp = Cv + R ==> Cp = 5R/2 and Cv = 3R/2 (similarly calculate others).
Cp or Cv for a mixture can be calulated using weighted average. (The proof of this is simple and is done in detail in Chemical thermodynamics).
Hence Cp of mixture = [ 1 x 5R/2 + 1 x 7R/2 + 1 x 8R/2 ] / 1 + 1 + 1 = 10R/3
Cv of the mixture = [ 1 x 3R/2 + 1 x 5R/2 + 1 x 3R ] / 1 + 1 + 1 = 7R/3
Hence Gamma of mixture = 10/7 = 1.428.
113rd one to be done as 2nd one, we have to find Cv,which comes out to be 4 cal
Ans 4) T is equal to 293K at NTP (20oC is normal room temperature). " The product of the gas constant R,i.e, 0.0821 lit atm/K/mole and NTP temperature in kelvin equals 22.4" This is not true the value is 22.4 at STP (T = 273 K). Z value remains constant for an ideal gas. (V value chages as T changes to keep Z constant).