Pressure of the gas in the tank at its melting point
T₁ = 298 K; P₁ = 2.98 atm; T₂ = 1100 K; P₂ = ?
\(\frac { { P }_{ 1 } }{ { T }_{ 1 } } =\frac { { P }_{ 2 } }{ { T }_{ 2 } } \)
\(\Rightarrow \quad { P }_{ 2 }=\frac { { P }_{ 1 } }{ { T }_{ 1 } } \times { T }_{ 2 }\)
\({ P }_{ 2 }=\frac { 2.98\quad atm }{ 298\ K } \times 1100\ K=11\ atm\)
At 1100 K the pressure of the gas inside the tank will become 11 atm. Given that tank can withstand a maximum pressure of 12 atm, the tank will start melting first. 

\(\frac { { \gamma }_{ unknown } }{ \gamma { N }_{ 2 } } =\frac { { t }_{ { N }_{ 2 } } }{ { t }_{ unknown } } =\sqrt { \frac { { m }_{ { N }_{ 2 } } }{ { m }_{ unknown } } } \)
\(\frac { 84\quad sec }{ 192\quad sec } =\sqrt { \frac { 14g\quad { mol }^{ -1 } }{ { m }_{ unknown } } } =\frac { 14g\quad { mol }^{ -1 } }{ { m }_{ unknown } } \)
\({ m }_{ unknown }=28 g { mol }^{ -1 }\times { \left( \frac { 192\quad sec }{ 84\quad sec } \right) }^{ 2 }\)
m_unknown = 146 g mol^-1 . 

Given,
V = 154.4 \(\times\) 10^-3 dm³,
P = 742 mm of Hg
T = 298 K m = ?
\(n=\frac { PV }{ RT } =\frac { 742mm\ Hg\times 154.4\times { 10 }^{ -3 }L }{ 62mm\ Hg\ L{ K }^{ -1 }{ mol }^{ -1 }\times 298K } \)
\(n=\frac { Mass }{ Molar\ Mass } \)
mass = n \(\times\) Molar mass
= 0.006 \(\times\) 2.016
= 0.0121 g = 12.1 mg.

Rate of diffusion \(\text { (r) } \alpha \frac{1}{\sqrt{M}}\)
\(\mathrm{M}_{\mathrm{NH}_{3}}=17 ; \mathrm{M}_{\mathrm{HCl}}=36.5 \)
\(\therefore \mathrm{r}_{\mathrm{NH}_{3}}>\mathrm{r}_{\mathrm{HCl}} \)
Hence white fumes are first formed near HCL.

On heating incineration might take place due to the increase of pressure

Astronauts must wear space suits filled with air, whenever they leave a space craft and are exposed to the environment of space.
Dangers experienced on moon's space :
1. In space there is no air to breathe and no air pressure.
2. Space is extremely cold and filled with dangerous radiations.
3. If they do not wear space suits, there may be bleeding due to high body pressure.
4. The space suits prevent astronauts from impacts of small bits of space dust.
5. There is no atmosphere on the moon and there are dangers from micro-meteorite impacts.
6. The astronauts are protected from these dangers on wearing space suits.

The van der Waals equation for n moles is
\(\left( P+\frac { { an }^{ 2 } }{ { v }^{ 2 } } \right) \left( v-nb \right) =nRT\) ....(1)
For 1 mole
\(\left( P+\frac { { a }^{ 2 } }{ { v }^{ 2 } } \right) \left( v-b \right) =RT\) ...(2)
From the equation we can derive the values of critical constants P_c, V_c and T_c, in terms of a and b, the van der Waals constants, On expanding the above equation
\(pv+\frac { a }{ v } -pb-\frac { ab }{ { v }^{ 2 } } -RT=0\)    ...(3)
Multiply equation (3) by V² / P
\(\frac { { v }^{ 2 } }{ p } \left( Pv+\frac { a }{ v } -pb-\frac { ab }{ { v }^{ 2 } } -RT \right) =0\)
\({ v }^{ 3 }+\frac { av }{ p } +-{ bv }^{ 2 }-\frac { ab }{ { v }^{ 2 } } -\frac { { RTV }^{ 2 } }{ p } \)    ...(4)
When the above equation is rearranged in powers of Y.
\({ v }^{ 3 }-\left[ \frac { RT }{ P } +b \right] { v }^{ 2 }+\left[ \frac { a }{ p } \right] v-\left[ \frac { ab }{ p } \right] =0\)        ...(5)
The equation (5) is a cubic equation in V. On solving this equation,
we will get three solutions. At the critical point all these three solutions of V are equal to the critical volume V_C. The pressure and temperature becomes Pc and T_c respectively
i.e., V = V_c
V - V_c = 0
(V - V_C)³ = 0
V3 - 3V_CV² + 3V_c²V - V_c³ = 0 .....(6)
As equation (5) is identical with equation (6), we can equate the coefficients of V², V and constant terms in (5) and (6).
\(-3{ v }_{ c }{ v }^{ 2 }=-\left[ \frac { { RT }_{ c } }{ { p }_{ c } } +b \right] { v }^{ 2 }\)
\(3{ v }_{ c }=\frac { { RT }_{ c } }{ { p }_{ c } } +b\)           .....(7)
\(3{ v }_{ c }^{ 2 }=\frac { a }{ { p }_{ c } } \)           ......(8)
\(3{ v }_{ c }^{ 2 }=\frac { ab }{ { p }_{ c } } \)           ...(9)
Divide equation (9) by equation (8)
\(\frac { { v }_{ c }^{ 3 } }{ 3{ v }_{ c }^{ 2 } } =\frac { ab/{ p }_{ c } }{ a/{ p }_{ c } } \)
\(\frac { { V }_{ c } }{ 3 } =b\)
i.e. V_c = 3b          .....(10)
when equation (10) is substituted in (8)
\(3{ v }_{ c }^{ 2 }=\frac { a }{ { p }_{ c } } \)
\({ p }_{ c }=\frac { a }{ 3{ v }_{ c }^{ 2 } } =\frac { a }{ 3\left( { 3b }^{ 2 } \right) } =\frac { a }{ { 3\times 9b }^{ 2 } } =\frac { a }{ { 27 }b^{ 2 } } \)
\({ p }_{ c }=\frac { a }{ { 27 }b^{ 2 } } \)            ...(11)
substituting the values of V_c and P_c in equation (7),
\(3vc=b+\frac { { RT }_{ c } }{ p } \)
\(3\left( 3b \right) =b+\frac { { RT }_{ c } }{ \left( \frac { a }{ { 27b }^{ 2 } } \right) } \)
\(9b-b=\left( \frac { { RT }_{ c } }{ a } \right) 27{ b }^{ 2 }\)
\(8b=\frac { { t }_{ c }R{ 27b }^{ 2 } }{ a } \)
\(\therefore { T }_{ c }=\frac { 8ab }{ 27R{ b }^{ 2 } } =\frac { 8a }{ 27Rb } \)
\({ T }_{ c }=\frac { { 8 }_{ a } }{ 27Rb } \)           ......(12)
The critical constants can be calculated using the values of van der Waals constant of a gas and vice versa.
\(a=3{ V }_{ C }^{ 2 }{ P }_{ C }\quad and\quad b=\frac { { V }_{ C } }{ 3 } \)