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Hint: Density of a gas is the ratio of its total mass to the volume that it occupies. Since, in this question it is not given that the gas is behaving like a real gas, we can assume that the gas is behaving ideally and use the ideal gas equation.

Complete step by step answer:

Density of a substance is the ratio of its mass to its volume. Here we will assume that the gas is an ideal gas. Since for ideal gases:

PV=nRT

\[PV=\cfrac { m }{ M } RT\]

Where ‘m’ is the total mass of the gas and ‘M’ is the molar mass of the gas.

\[ \Rightarrow PM=\cfrac { m }{ V } RT\]

\[\Rightarrow \rho =\cfrac { P\times M }{ R\times T } \]

Where ‘$\rho $’ is the density of the gas.

Therefore, $ \rho =\cfrac { 2\quad bar\times M }{ R\times 300K } $ (Here the temperature is 27$^{ o }{ C }$ which is equal to 300 K)…………………………………………………………………………………………………….(1)

Similarly, the same equation at STP condition will be:

\[{ \rho }_{ STP }=\cfrac { 1\quad bar\times M }{ R\times 273K } \]………………………………………………(2)

Dividing equation (2) by (1) we get:

\[\cfrac { { \rho }_{ STP } }{ \rho } =\cfrac { \cfrac { 1\quad bar\times M }{ R\times 273K } }{ \cfrac { 2\quad bar\times M }{ R\times 273K } } \]

\[\Rightarrow \cfrac { { \rho }_{ STP } }{ 5.46g{ L }^{ -1 } } =\cfrac { 1\quad bar\times 300K }{ 2\quad bar\times 273K } \]

Therefore,

\[\Rightarrow { \rho }_{ STP }=\cfrac { 1\quad bar\times 300K }{ 2\quad bar\times 273K } \times 5.36g{ L }^{ -1 }\]

\[\Rightarrow { \rho }_{ STP }=3.00g/{ dm }^{ 3 }\]

Note: Do not get confused between STP and NTP conditions. In STP (standard temperature and pressure) conditions, the temperature is 0$^{ o }{ C }$ and the pressure is 1 bar. In NTP (normal temperature and pressure) conditions, the temperature is 20$^{ o }{ C }$ and the pressure is 101.325 kPa.

Complete step by step answer:

Density of a substance is the ratio of its mass to its volume. Here we will assume that the gas is an ideal gas. Since for ideal gases:

PV=nRT

\[PV=\cfrac { m }{ M } RT\]

Where ‘m’ is the total mass of the gas and ‘M’ is the molar mass of the gas.

\[ \Rightarrow PM=\cfrac { m }{ V } RT\]

\[\Rightarrow \rho =\cfrac { P\times M }{ R\times T } \]

Where ‘$\rho $’ is the density of the gas.

Therefore, $ \rho =\cfrac { 2\quad bar\times M }{ R\times 300K } $ (Here the temperature is 27$^{ o }{ C }$ which is equal to 300 K)…………………………………………………………………………………………………….(1)

Similarly, the same equation at STP condition will be:

\[{ \rho }_{ STP }=\cfrac { 1\quad bar\times M }{ R\times 273K } \]………………………………………………(2)

Dividing equation (2) by (1) we get:

\[\cfrac { { \rho }_{ STP } }{ \rho } =\cfrac { \cfrac { 1\quad bar\times M }{ R\times 273K } }{ \cfrac { 2\quad bar\times M }{ R\times 273K } } \]

\[\Rightarrow \cfrac { { \rho }_{ STP } }{ 5.46g{ L }^{ -1 } } =\cfrac { 1\quad bar\times 300K }{ 2\quad bar\times 273K } \]

Therefore,

\[\Rightarrow { \rho }_{ STP }=\cfrac { 1\quad bar\times 300K }{ 2\quad bar\times 273K } \times 5.36g{ L }^{ -1 }\]

\[\Rightarrow { \rho }_{ STP }=3.00g/{ dm }^{ 3 }\]

__Hence the density of gas at STP conditions is $3.00g/{ dm }^{ 3 }$.__Note: Do not get confused between STP and NTP conditions. In STP (standard temperature and pressure) conditions, the temperature is 0$^{ o }{ C }$ and the pressure is 1 bar. In NTP (normal temperature and pressure) conditions, the temperature is 20$^{ o }{ C }$ and the pressure is 101.325 kPa.

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