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. 

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.

V₁ = 8.5 dm³ V₂ = 6.37 dm³
T₁ = ? T₂ = 0^o C = 273 K
\(\frac { { V }_{ 1 } }{ { T }_{ 1 } } =\frac { { V }_{ 2 } }{ { T }_{ 2 } } \)
\({ V }_{ 1 }\times \left( \frac { { T }_{ 2 } }{ { V }_{ 2 } } \right) ={ T }_{ 1 }\)

T₁ = 364.28 K

It will be more difficult to drink water with a straw on the top of a Mount Everest. This is because the reduced atmospheric pressure is less effective in pushing water up into the straw, below the water surface.
The force that propels water through a straw is atmospheric pressure, which is less at high altitude.

n =1.82 mole
V = 5.43 dm³
T = 69.5 + 273 = 342.5
P = ?
PV = nRT
\(P=\frac { nRT }{ V } \)

P = 9.425 atm.