Answer
No, it would exert a pressure of 11 bar
Work Step by Step
Molar mass $(Ar)$ : $40 \space g/mol$
Perfect gas law: $pV = nRT$
R in $dm^3 \space bar \space K^{-1} \space mol^{-1} = 8.31447 \times 10^{-2}$
$$pV = nRT \longrightarrow p = \frac{nRT}{V}$$ $$n =\frac{m}{mm} = \frac{25 \space g}{40 \space g/mol} = 0.625 \space mol$$ $$T(K) = T(^oC) + 273.15 = 30 + 273.15 = 303.15 $$
$$p = \frac{(0.625 \space mol)( 8.31447 \times 10^{-2} \space dm^3 \space bar \space K^{-1} \space mol^{-1} )(303.15 \space K)}{1.5 \space dm^3}$$ $$p = 11 \space bar$$
