If the volume of a given mass of gas at 298 K and pressure of 205.2 x 10\(^3\) Nm\(^{-2}\) is 2.12dm\(^3\), what is volume of the gas at s.t.p? (Standard pressure = 101.3 x 10\(^3\) Nm\(^{-2}\), Standard temperature = 273K)
- 39.3dm 3
- 22.4dm3
- 4.93dm3
-
3.93dm3
- 0.393dm3
The correct answer is: D
Explanation
V\(_1\) = 2.12dm\(^3\), V\(_2\) = ?
P\(_1\) = 205.2 x 10\(^3\) Nm\(^{-2}\), P\(_2\) = 101.3 x 10\(^3\) Nm\(^{-2}\)
T\(_1\) = 293K, T\(_2\) = 273K
Using \(\frac{P_1 * V_1}{T_1}\) = \(\frac{P_2 * V_2}{T_2}\)
: V\(_2\) = \(\frac{P_1 * V_1 * T_2}{P_2 * T_1}\)
→ \(\frac{205.2 x 10^3 * 2.12 * 273}{101.3 x 10^3 * 293}\)
= 4.0dm\(^3\)