Question

look at the attachement and asxnwer the questions

Answer 1

1.  P1  V1 / T1 =  P2   V2 / T2   - Ideal gas law
      V2 =   P1  V1  T2 / (T1  P2)
              =  380 mm * 1 Litre * 288 K / [ 303 K  * 750 mm ]

2.    P V = n R T          ideal gas law
      n = R T / (P  V )  =  8.314 * (273+550) / [ 0.2 * 1.013 *10^5  * 0.350)
         n =  m / M
       =>          M =  m / n  =   0.05 / n
 
3.    P V = n R T = m R T / M
       M =  m R  T / [ P V ]  = d  R T / P,    d = density
            = 1.5 * 10^-3 * 8.314 * (273+65) / [750 * 1.013 * 10^5 / 760 ]
            = 1.5 * 10^-3 * 8.314 * (273+65) * 760 / [750 * 1.013 * 10^5  ]

4. 
which conditions ?  where are they?
5.
open vessel   pressure P and V volume of gas remain constant.
T1 = 300 K.     P1=P ,      V1 = V,     n1 = moles present
               P  V = n1 R T1
P    and V2 = V,        n2   <  n1          T2 > T1
            P  V =  n2  R  T2,          n2 moles of gas remains after heating
         n2 T2 = n1 T1
          n2 =  (1-3/5) n1 = 2/5 * n1
           =>   T2 / T1 = n1/n2 = 5/2 = 2.5
         T2 = 300K * 2.5 = 750 K  ,   at this temperature only 2/5th of initial gas is present.

a)  If the vessel now is heated to  T3 = 900 K, then
         n3 / n1 = T1 / T3 = 1/3           =>  only 1/3 is present.  and  2/3 of air escapes out.
b)  let       n4 /n1 = 1/2
        n4/n1 = T1/ T4     =>   T4 = 600K
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6.
V = 10L =  0.01 m^2        T = 300K
p  = m R T / (MV)
p_He = 0.4  8.314 * 300 / (4 * 0.01 )     Pa
p_O2 = 1.6 * 8.314 * 300 / (32 * 0.01)   Pa
p_N2  = 1.4 * 8.314 * 300 / (28 * 0.01)  Pa
add the three. to get total pressure.
P = 8.314 * 300 [ 0.4/4 + 1.6 /32 + 1.4/28 ] / 0.01 
====================
7.   one of N2 gas =  N = avogadro num = 6.023 * 10^23 number of molecules.
 volume of each molecule =  4/3   Pi  r^3,        r = 2 * 10^-10 m
volume of gas at NTP,  22.4 litres = 0.0224    m^3
empty space =      0.0224  -  6.023 * 10^23 * 4/3  pi   2^3   10^-30 

8.
P_A = 2 P_B              V_A = 2 V_B               T_A = 2 T_B
P V / RT = n   =>      n_A  / n_B  =   2
     =>   m_A = 2  m_B
===========
9.
m1 = 5.40 gm        T1 = 300 K              V1 = V           P1 = P
m2 = 0.14 gm     M2 = 2        n2 = 0.07            T2 = 290 K          V2 = V         P2 = P
n1 T1  =  n2  T2     =>    n1 = 0.07 * 290 / 300
    M =   m1 / n1
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10.          P ,    d = m / V1         T1 = 300 K
            P =  d R T / M        =>         d1 T1 = d2 T2        
                 0.75 d / d =  300 / T2          => 400 K
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11

V = 1 litre       P = 1 atm     T = 373 K
num of moles in steam:      n2 = P V / R T  = 3.381 * 10^-6

 mass of this 1 litre of vapour  at 373 K,   m2 = density * Volume
        =>  m2 = 0.0006 g/cc * 1000 cc =  0.6 gms

In liquid form,   the water molecules weighing  0.6 gms  occupy  0.6 cc.
If we assume that in water , the molecules are very closely packed (as compared to the gas vapor), then    the volume occupied by water molecules is 0.6 cc   in one litre (1000 cc) of the water vapour  at 1 atm at 373 K.