The pressure exerted by 12 g of an ideal gas at temperature `T(""^(@)C)` in a vessel of volume V is 1 atm. When the temperature is increased by `10^(@)C` at the same volume, the pressure increases by 10%. If molecular weight of the gas is 120, the temperature `(T^(@)C)` and volume (V) are

Options

A. `T = -273^(@)C , V = 0.082 L`
B. `T = -173 ^(@)C , V = 0.82 L`
C. `T = 0^(@) C , V = 22.4 L`
D. `T = 27^(@) C , V = 22.4 L`

Answer

B

Explanation

Moles of gas `(12)/(120) = 0.1`
pV = nRT or `1 xx V = 0.1 xx R (273 + t) " " …. (1)`
After temperature change , `p = 1 + (10)/(100) =1 .1 ` atm
`T = (t + 273 + 10) = (283 + t) K " " … (2)`
Dividing Eq. (2) byh Eq. (1) `(1.1)/(1) = (283 + t)/(273 + t) implies t = - 173^(@) C`
Substituting the value of t in Eq. (2) , `1.1 xx V = 1 (283 - 173) xx 0.082 implies V = 0.82 L`