A 92.3 g sample of O2 gas at 0.0°C and 5.065 x 104 Pa is compressed and heated until the volume is 2.76 L and the temperature is 27°C?

PV = nRT  and  n = m/M
Then, PV = (m/M)RT
Hence, P = mRT/(VM)
In the final conditions: P₂ = mRT₂/(V₂M)

Mass, m = 92.3 g
Gas constant, R = 8.314 J / (mol K)
Temperature, T = (273 + 27) K = 300 K
Final volume, V₂ = 2.76 L = 0.00276 m³
Molar mass, M = 16.0 × 2 g/mol = 32.0 g/mol

Final pressure, P₂ = 92.3 × 8.314 × 300 / (0.00276 × 32.0) Pa = 2.61 × 10⁻⁶ Pa

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Remarks: The initial temperature and initial pressure may be redundant.

V = nRT / P = (92.3 g O2 / (31.99886 g O2/mol)) x (8.3144626 L kPa/K mol) x
(273.15 K) / (50.65 kPa) = 129.3 L

(5.065 x 10^4 Pa) x (27 + 273) K / (0.0 + 273) K x (129.3 L / 2.76 L) = 2607519 Pa =
2.61 x 10^6 Pa

P * V = n * R * T

P[1] * V[1] = n * R * T[1]
P[2] * V[2] = n * R * T[2]

P[1] * V[1] / T[1] = n * R
P[2] * V[2] / T[2] = n * R

P[1] * V[1] / T[1] = P[2] * V[2] / T[2]

P[1] = 5.065 * 10^4 Pa
V[1] = ?
T[1] = 0 C = 273.15 K
P[2] = ?
V[2] = 2.76 L
T[2] = 27 C = 300.15 K

5.065 * 10^4 * V[1] / 273.15 = P[2] * 2.76 / 300.15

We need to find V[1]

5.065 * 10^4 * V[1] = n * R * 273.15

O2 is 32 grams per mole

92.3 g / (32 g/mol) = (92.3/32) mol

V[1] = (92.3 / 32  mol) * R * 273.15 K / (5.065 * 10^4 Pa)
V[1] = (92.3 * 273.15 / (32 * 5.065 * 10^4)) * R [mol * K / Pa]

We need a value for R that is in terms of Pa * L / (mol * K)

https://en.wikipedia.org/wiki/Gas_constant

R = 8.31446261815324 * 10^3 Pa * L / (mol * K)

V[1] = 92.3 * 273.15 * 8.31446261815324 * 10^3 / (32 * 5.065 * 10^4) Liters

5.065 * 10^4 * V[1] / 273.15 = P[2] * 2.76 / 300.15
5.065 * 10^4 * 300.15 * V[1] / (273.15 * 2.76) = P[2]

P =>
5.065 * 10^4 * 300.15 * (92.3 * 273.15 * 8.31446261815324 * 10^3) / (32 * 5.065 * 10^4 * 273.15 * 2.76) =>
300.15 * 92.3 * 273.15 * 8.31446261815324 * 10^3 / (32 * 273.15 * 2.76) =>
300.15 * 92.3 * 8.31446261815324 * 10^3 / (32 * 2.76) =>
2,608.0455574231379892187... * 10^3 =>
2.608.... * 10^6

2.608 * 10^6 Pascals