A container of fixed volume 0.4 L contains 0.56 g of a gas at `27^(@)C`. The pressure of the gas at this temperature is 936 mmHg. If the amount of the gas is increased to 2.1g and its temperature is decreased to `17^(@)C`, then what will be the pressure of the gas? Assuming gas behaves ideally.

Explanation

Suppose, the molarr mass of the gas=`Mg*mol^(-1)`
so, 0.56 of the gas`=(0.56)/(M)` mol of the gas
At the initial state, `P=(936)/(760)=1.23 atm, n=(0.56)/(M)mol`,
V=0.4L,
And T=(273+27)K=300K
Substituting the values of P, N, V and T in ideal gas equation, PV=nRT, we have ltBrgt `1.23xx0.4=(0.56)/(M)xx0.0821xx300`
`therefore`molar mass of the gas=`28g*mol^(-1)`
At the second condition: `V=0.4L,n=(2.1)/(28)=0.075mol`,
`T=(273+17)=290K`
Again, applying the equation, PV=nRT, we have
`Pxx0.4=0.075xx0.0821xx290`
`therefore P=4.46atm`
So, the pressure of gas at the second condition is 4.46 atm.