From the first set of data, the molar mass of the gas can be calculated:

P = 101.3 kPa, T = 273 K, d = 1.375 g L^{-1}

Therefore, for V = 1.00 L, mass of gas = 1.375 g

Let MW be the molar mass of the gas

Then number of moles present = n = (1.375 / MW)

Substitute in the ideal gas equation:

101.3 x 1.00 = (1.375 / MW) x 8.313 x 273

Therefore MW = (1.375 x 8.314 x 273) / (101.3 x 1.00) = 30.8

Using this result, the mass (m) of 1.00 L of the gas at the new
conditions can be deduced by again using the IGE:

P = 98.0 kPa, T = 303 K, moles of gas, n, = (m / 30.8)

98.0 x 1.00 = (m / 30.8) x 8.314 x 303

m = (98.0 x 1.00 x 30.8) / (8.314 x 303) = 1.199 g

As this is the mass of gas in 1.00 L, then density = m / V = 1.199
g L^{-1}.