JEE Main 2023 · 1 February · evening · Single correct · +4 / −1
For three low density gases A, B, C pressure versus temperature graphs are plotted while keeping them at constant volume, as shown in the figure.
The temperature corresponding to the point 'K' is :
- (a) -273° C
- (b) -373° C
- (c) -100° C
- (d) -40° C
Answer and solution
Answer (a) -273° C
Solution
Physics sir · 4 steps
Every constant-volume line ends at zero pressure, and that happens only at absolute zero.
- Constant volume. For an ideal gas at fixed volume, PV = nRT gives P = nR/V T, where T is in kelvin. So pressure is directly proportional to absolute temperature.
P = nR/V (t + 273), t in °C
- Different slopes. Gases A, B and C have different values of n/V, so their lines have different slopes. But every line reaches P = 0 at the same place, t + 273 = 0.
- Point K. K is where all three lines meet the temperature axis, so P = 0 there. That point is T = 0 K, which is t = -273°C.
- So. The temperature at K is absolute zero, -273°C, which is option (a).
The idea to keep. Extend any P–T line at constant volume, or any V–T line at constant pressure, back to zero and it always lands at -273°C, whatever the gas or its amount.
Answer from the NTA final answer key. Solution by PrepWiser.