At a constant temperature of 27 degrees and a pressure of 10 ^ 5 Pa, the volume of gas is 1 m ^ 3.

At a constant temperature of 27 degrees and a pressure of 10 ^ 5 Pa, the volume of gas is 1 m ^ 3. At what temperature this gas will occupy a volume of 0.5m ^ 3. at the same pressure 10 ^ 5 Pa.

Given:
T1 = 27 C = 273 + 27 = 300 K
p1 = 10 ^ 5 Pa
V1 = 1 m ^ 3
p2 = 10 ^ 5 Pa
V2 = 0.5 m ^ 3
T2 -?
p1 * V1 = m1 * R1 * T1 / M1 – Cliperon-Mendeleev equation
p2 * V2 = m2 * R2 * T2 / M2
since the gas is the same, then m1 = m2 – the masses are the same, M1 = M2 – the molar masses are the same,
R1 = R2 = 8.31 J / mol * K – universal gas constant
m1 * R1 / M1 = p1 * V1 / T1
m2 * R2 / M2 = p2 * V2 / T2
m1 * R1 / M1 = m2 * R2 / M2 and therefore:
p1 * V1 / T1 = p2 * V2 / T2
T2 = p2 * V2 * T1 / p1 * V1
T2 = 10 ^ 5 * 0.5 * 300/10 ^ 5 * 1 = 150 K