Question

If the density of a certain gas at 30°C and 768 atm pressure is 1.35 kg/$m^{3}$, find its density at STP.

Answer 1

"We can use the formula PV = nRT

Let us assume that, V = 1 litre

PV = nRT

$n\quad =\quad \frac { PV }{ RT } \quad =\quad \frac { \left( \frac { 768 }{ 760 } \quad atm \right) \quad \times \quad 1L }{ \left( 0.0821\quad \sfrac { L\quad atm }{ mole\quad K }\right) \quad \times \quad \left( 303\quad K \right)} \quad =\quad 0.0406\quad moles\quad of\quad gas$

$=\quad 1L\quad \times \quad \left( \frac { 1 { m }^{ 3 } }{ 1000\quad L}\right) \quad \times \quad \left( \frac { 1.36\quad kg }{ 1\quad { m }^{ 3 } }\right) \quad \times \quad \left( \frac { 1000\quad g}{ 1\quad kg}\right) \quad =\quad 1.35\quad g$

Thus, MW of gas $=\quad \frac { 1.35\quad g }{ 0.0406\quad moles } \quad =\quad 33.3\quad \sfrac { g }{ mole }$

At STP, $n\quad =\quad \frac { PV }{ RT } \quad =\quad \frac { 1.0\quad atm\quad \times \quad 1\quad L}{ 0.0821\quad \sfrac { L\quad atm}{ mole\quad K} \quad \times \quad 273\quad K} \quad =\quad 0.0446\quad moles$

$=\quad 0.0446\quad moles\quad \times \quad \left( 3.33{g }/{ mol}\right) \quad \times \quad \left( \frac { 1\quad kg}{ 1000\quad g}\right) \quad =\quad 0.0149\quad kg$

$=\quad \frac { 0.00149\quad kg }{ 1\quad L\quad \times \quad \left( \frac {1000\quad L }{ 1\quad {m}^{3}}\right)} \quad =\quad 1.49\quad \sfrac {kg}{{m}^{3}}$"

Answer 2

Risk capital establishment established for:-

a) Import analysis
b) Export analysis
c) none
d) both